§^ ©hcowtical and §<:a(tiat Statist 



ON THE 



STRENGTH 



OF 



BEAMS AND COLUMNS; 

IX WHICH 

THE ULTIMATE A^B THE ELAS- 
TIC LIMIT STRENGTH OF BEAMS AND COL- 
UMNS IS COMPUTED FROM THE ULTIMATE AND ELASTIC 
LIMIT COMPRESSIVE AND TENSILE STRENGTH OF THE MATERIAL, 
BY MEANS OF FORMULAS DEDUCED FROM THE CORRECT 
AND NEW THEORY OF THE TRANSVERSE 
STRENGTH OF MATERIALS. 

BY / 

EOBEET H. COUSI]Srs, 

Civil Engineer, 

Formerly Assistant Pi'ofessor of Mathematics at the Virginia Military Institute, 

Lexington, Ya. 






E. & F. N. SPON, 



NEW YORK : 
12 CORTLANDT STREET. 



LONDON : 

12 5 Strand, 



1889, 



{All tights 7'eserved.] 






.t** 



Copyright, 1889, 

BY 

R. H. COUSINS. 

■ e B6 



INTEODUCTIOK. 



For more than two centuries the mathematical and me- 
chanical laws that govern the transverse strength of Beams and 
Columns have received the attention of the most expert 
mathematicians of all countries. Galileo, in 1()38, formulated 
and published the first theory on the subject. He was fol- 
lowed bj such philosophers as Mariotte, Leilmitz, Bernouilli, 
Coulomb, and others, each amending and extending the work 
of his predecessor, until the year 1824:, when Navier succinctly 
stated the theory that is recognized to be correct at the present 
day, and to which subsequent writers and investigators have 
added but little. 

This theory has neither received the endorsement of the 
experimenters nor of some of the theoretical writers. *' Ex- 
cepting as exhibiting approximately the laws of the phenom- 
ena, the theory of the strength of materials has many prac- 
tical defects" (Wiesbach). " It has long been known that under 
the existing theory of beams, which recognizes only two ele- 
ments of strength — namely, the resistance to direct compression 
and extension — the strength of a bar of iron subjected to a 
transverse strain cannot be reconciled with the results obtained 
from experiments on direct tension, if the neutral axis is in 
the centre of the bar" (Barlow). 

During the present century much time and means have 
been expended in attempts to solve, experimentally, the prob- 
lems that have eno^aofed the attention of tlie mathematicians, 
and as the result of their labors we find such experimenters as 
Ilodgkinson, Fairbairn, and others, whose names are house- 



IV INTRODUCTION 

hold words in tlie literature of the suhject, adopting empirical 
rules for the strength of Beams and Columns rather than the 
rational formulas deduced bj the scientists. " For no theory 
of the rupture of a simple beam has jet been proposed which 
fully satisfies the critical experimenter" (De Yolson Wood). 

That we should be able to deduce the strength of Beams and 
Columns from the known teiisile and compressive strength of 
the material composing them, has been apparent to many 
writers and experimenters on the subject, but to the present 
time no theory has been advanced that embodies the mathe- 
matical and mechanical principles necessary to its accomplish- 
ment. The theory herein advanced and the formulas resulting 
therefrom deduce the strength of Beams and Colunms from 
tlie direct crushing and tensile strength of the material com- 
posing them, without the aid of that coefficient that has no 
place in ^nature, the Modulus of Rupture. The theory and 
the formulas deduced therefrom are in strict accord with cor- 
rect mechanical and mathematical principles, and the writer 
believes that they will fully satisfy the results obtained by the 
experimenter. 

The great practical benefits to be derived from the correct 
theory of the strength of Beams and Columns will be evident, 
when we consider the countless tons of metal that have 
been made into railroad rails, rolled beams, and the other 
various shapes, and that the manufacturers were without 
knowledge of the work to be performed by the diiferent parts 
of the beam or column in sustaining the load that it was in- 
tended to carry. The best that they have been able to do is 
to compute the strength by the aid of an empirical quantity 
deduced from experiments on '' similar beams." The correct 
theory will enable them to foretell the strength of any untried 
shape, and the reason for the strength of those that have been 
long m use, which is the " true object of theory." 

R 11 

Dallas, Texas, March 13, 1889. * * " 



CONTENTS. 



CHAPTER I. 



1. 

2. 

3. 

4 

5. 

6. 

7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 
16. 
17. 



FORCES DEFINED AND CLASSED. 

Force Defined, 

Stress or Strain, 

The Load, 

Equilibrium and Resultant 

Bending Moment— Concentrated Forces, 
General Formulas for Bending Moments, 
Uniformly Varying Forces— Rectangular Areas, 
Resultant, ....... 

Moment of Uniformly Varying Forces, 



Uniformly Varying Forces — Circular Segment Areas, 

Moment of Uniformly Varying Forces — Circular Segment Areas, 

Resultant, 

Moment of Uniformly Varying Forces — Circular Arcs, 

Resultant, " " " . " " . . 



3 
6 

8 
8 
9 
11 
13 
13 
14 
16 
17 
18 
19 



CHAPTER II. 

RESISTANCE OF CROSS SECTIONS TO RUPTURE. 

18. Moment of Resistance, ...... 

19. Neutral Line, 

20. Bending A and Moment of Resistance, . 

21. Equilibrium, 

22. Position of the Neutral Line, ..... 

23. Neutral Line at the Transverse Elastic Limit, 

24. Movement of the Neutral Line, with the Deflection, 

25. Neutral Lines of Rupture in a Rectangular Section, 

26. Relative Value of the Compressive and Tensile Strains, 

27. Summary of the Tlieory, 



4il 

. 21 


22 


. 23 


24 


. 24 


26 


. 28 


s, . 28 


. 31 



vi CONTENTS. 

CHAPTER III. 

TRANSVERSE STRENGTH. 

ART, PAGE 

28. Coefficients of Strength, 38 

29. Elasticity of Materials, 38 

30. Elastic Limits, 34 

31. Elastic Limit of Beams, 34 

32. Working Load and Factors of Safety, .34 

33. General Formula — Transverse Strength, 35 

34. Relative Transverse Strength of a Beam, 35 

35. Moment of Resistance — Rectangular Sections, .... 37 

36. The Xeutral Line " " .... 38 

37. Transverse Strength " " .... 38 

38. Designing a Beam " " .... 39 

39. To Compute the Compressive Strain, 39 

39. " " " Tensile " 39 

40. Moment of Resistance — Hodgkinson Sections, . . . .40 
4L Neutral Line " " .... 42 

42. Transverse Strength " " .... 44 

43. To Design a Hodgkinson Section, 44 

44. Moment of Resistance — Dbl. T and Hollow Rectangular Sections, 47 

45. Neutral Line " " " " " 48 

46. Transverse Strength .* «< << " .< 4^ 

47. To Design a " " " " " 48 

48. Moment of Resistance — Invtd T, Dbl. Invtd. T and U Sections, 52 

49. Neutral Line " " " " " " 58 

50. Transverse Strength " " " " " " 53 

51. To Design an " " " " " " 53 

52. Moment of Resistance — Circular Sections, . . . . .55 

53. Neutral Line. " " .... 56 

54. Transverse Strength " " 56 

55. To Compute the Compressive Strain of Circular Sections, . 58 

56. " " " Tensile .<.<.< << ^ 5^ 

57. Relative Strength of Circular and Square Sections, . . . 59 

58. Moment of Resistance — Hollow Circular Sections, . . .60 

59. Neutral Line " " " ... 60 

60. Transverse Strength " " " ... 60 

CHAPTER IV. 

CAST- 1 RON BEAMS. 

61. Compressive Strength of Cast-Iron, ...,,. 64 

62. Tensile " " " 64 



CONTENTS. VU 

ABT. PAGE 

68. Ratio of the Compressive to tlie Tensile Strength, ... 65 

64. Transverse Strength of Cast-iron, 65 

65. To Compute the Compressive Strength of Cast-Iron, . . 67 

65. " " " Tensile ....... ... 67 

66. Neutral Line— Rectangular Cast-Iron Beams, .... 71 

67. Transverse Strength " *' " .... 71 

68. To Design a Rectangular Cast-iron Beam, .... 73 

69. Hodgkinson Beams, 74 

70. Mr. Hodgkinson 's Experiments, 74 

71. Neutral Line — Hodgkinson Cast-Iron Beams, . . . .75 

72. Transverse Strength " " " . , . . 75 

73. To Design a Hodgkinson Cast-Iron Beam, 78 

74. Neutral Line — Double T and Box Cast-Iron Beams, . . 78 

75. Transverse Strength " " " " " . . .78 

76. To Design a Double T and Box Cast-Iron Beam, ... 79 

77. Neutral Line— Cast-Iron Circular Beams, 80 

78. Transverse Strength " " *' .... 81 

79. Movement of the Neutral Line 82 

80. Relative Strength of Circular and Square Cast-Iron Beams, . 83 

81. Neutral Line — Hollow Circular Cast-Iron Beams, . . .85. 

82. Transverse Strength " " " " . . . 86 

CHAPTER V. 

WROUGHT-IRON AND STEEL BEAMS. 

88. Compressive Strength of Wrought-Iron , . . . . 87 

84. Tensile .... .. g;. 

85. Compressive " " Steel, 88 

86. Tensile ...... gg 

87. To Compute Compressive Strength of Wrought-Iron and Steel, 88 

88. " " Tensile " " " " " 90 

89. Transverse Strength of Wrought-Iron Beams, .... 91 

90. " " " Steel Beams, 92 

91. Neutral Line — Rectangular W^rought-Iron and Steel Beams, . 92 

92. Transverse Strength " " - .« «« , 92 
98. Neutral Line— W^rt. Iron and Sfl Dbl. T and Rolled Eye-Beams, 96 

94. Transverse Strength " " " " " " " " 96 

95. Transverse Strength — Wrought-Iron and Steel Double T Beams, 

Flanges Unequ<al, ......... 96 

96. Transverse Strength — Wrought-Iron and Steel Double T Beams, 

Flanges Equal 98 

97. Neutral Line — Circular W^rought-Iron and Steel Beams, . . 100 

98. Transverse Stren^jth " " ' . 101 



Vlll CONTENTS. 

ART. PAGE 

99. Neutral Line— Hollow Circular Wrought-Iron Steel Beams, . 104 

100. Transverse Strength " " " " " . 104 

'CHAPTER VI. 

TIMBER BEAMS. 

101. The Compressive and Tensile Strength of Timber, . . . 106 

102. To Compute the Compressive and Tensile Strength of Timber, 106 

103. Neutral Line— Rectangular Wooden Beams, . . " . . 107 

104. Transverse Strength " " " .... 108 

105. Neutral Line— Circular " " . . . .111 

106. Transverse Strength " " " . . . . 113 

107. Relative Strength of Square and Circular Timber Beams, . .113 

CHAPTER Vn. 

STRENGTH OF COLUMNS. 

lOS. General Conditions of Failure of Columns, . . . . 114 

109. Resistance of the Cross-Section of the Column 119 

110. Notation, 121 

111. Deflection of Columns, . 121 

112. Classification of Columns, 123 

113. Columns that fail with the full Crushing Strength of the Material, 125 

114. Columns that fail with less than the full Crushing Strength of 

the Material and without Deflection, 125 

115. Columns that fail with less than the full Crushing Strength of the 

Material and with Deflection, . . . . . . .127 

116. Columns that Cross-break from Compression, . . . 130 

117. Columns that Cross- break from Compression and Cantileverage, 137 

CHAPTER YKL 

COMBINED BEAMS AND COLUMNS. 

118. General Statement, 145 

119. Notation, 146 

INCLINED BEAMS. 

120. General Conditions, 147 

121. Inclined Beam Fixed and Supported at one end, . . . 148 

Case I. — When the Load is applied at the free end of the Beam. 

Case II. — When the Load is uniformly distributed over the un- 
supported length of the Beam. 



CONTENTS. IX 

ART. PAGE 

123, Inclined Beam supported at one end and stayed or lield in po- 
sition at the other without vertical support, . . . .151 

Case I. — When tlie Inclined Beam is loaded at its stayed end. 
Case II. — When the Inclined Beam is loaded at its middle. 
Case III. — When the Load is uniformly distributed over the 
length of the Inclined Beam. 

Case IV. — When the Inclined Beam is loaded at its middle, and 

with an additional Load at its stayed end. 
Case V. — When the Load is uniformly distributed over the length 

of the Inclined Beam, and an additional Load applied to its 

stayed end. 

123. Inclined Beam supported vertically at both ends and loaded at 

its middle, . . . . \ 160 

124. Inclined Beam supported at both ends and the Load uniformly 

distributed over its length, 168 

trussed beams. 

125. General Conditions, 164 

Kot£. — The writer regrets that chapters on the following subjects are un- 
avoidably omitted at this time, " The Strength of Arches" and " The De- 
flection of Beams," as they were not satisfactorily complete, and "Beams 
of Maximum Strength with Minimum Material," as he was not fully pro- 
tected by letters-patent at the time of going to press. 



STKEIS^GTH OF BEAMS AND COLUMlSrS. 



CHAPTER 1. 

FORCES. 
Section 1.— Forces Defined and Classed. 

1. Force defined. Force may be defined to be an 
action between two bodies, or parts of a body, which eitlier 
causes or tends to cause a change in their condition, w^hether 
of rest or of motion ; and our knowledge of a force is prac- 
tically complete when we know its intensity^ expressed in 
pounds or some other imit of measure, its direction^ whether 
toward or from the body upon which it acts, its point of 
application^ and the angle that its line of direction makes 
with the surface of the body. 

For an unit of measure of forces we shall use the pound 
avoirdupois, as all of our recorded knowledge of the strength 
of materials used in structures is expressed in pounds on the 
square inch. 

2, Stress and Strain are words used to define that class 
of forces tliat are brought into action when contiguous parts 
of a body are caused to react upon each other by reason of 
the application of other forces to the body, and may be classed 
as follows : 

Oo7npression, Thrtist, or Pressure is the force exerted when 
the contiguous parts of a body are caused to move toward 
each other. 

Te?ision, Pidl, or Tensile Strain is the force exerted when 



2 STEENGTH OF BEAMS AND COLUMN'S. 

the coTitigiioiis parts of a body are caused to move away from 
each other. 

3. The Load. The external forces apphed to a body to 
produce the various kinds of stress or strain is called the 
Load, and its amount or magnitude is expressed in pounds ; it 
may be classed as follows : 

Concentrated Forces or Loads. While in nature every force 
must be distributed over a definite amount of surface, it is 
necessary, in order to define certain principles, to consider 
them to be concentrated at a single point, the effect being 
identical with that of the distributed load. 

A Distrihuted Force of uniform intensity is the force or 
load that acts with the same intensity on each square inch of 
the surface of the body over whicli it is distributed. 

A Distrihuted Farce of uniformly varying intensity is a 
force that increases in intensity in direct proportion to the 
distance from a given point. 

4. Equilibrium and Resultant. Fqnilihrium of a 
system of forces is such a condition that the combined action 
of the forces produces no change in the rest or motion of the 
body to which it is applied. 

The Liesultant of a system of forces is a single concen- 
trated force that will produce the same effect upon the body, 
if applied, that the system of forces will produce. 



CONCENTRATED PARALLEL FORCES. 



3 



Section II. — Concentrated Parallel Forces. 

5. Bending Moment. In this section will be deduced 
the relation existing between the vertical forces caused by 
gravity, the supporting forces and the horizontal distances 
between their lines of action. In order to ascertain the rela- 
tion between such forces it is necessary to obtain the product 
of each force by the perpendicular distance of its line of 
action from a given point. Such product is called the Bending 
Moment of the force. 

The vertical forces will be called the loads, and the connec- 
tion between their lines of action will be called a beam, with- 
out reference to the shape of its cross-section, its material, or 
its ability to resist the hending moment of the applied loads, 
which will be considered in the sequel. 

The following notation will be used throughout : 

L = the total applied load in pounds. 

,9 = the S2?an, the horizontal distance between the supports 
in inches. 

3f =: the hending moment. 

Case I. — The Beam fixed at one end and loaded at the 

OTHER. 

A load thus applied will hend the beam (Fig. 1) at each 




section from the free end to the point of support, but un- 
equally. The bending moment at any section is the product of 



4 STRENGTH OF BEAMS AND COLUMNS. 

the load by the distance of the section from the free end of 

the beam. The Greatest Bending Moment, at the ^point of 

support, will be 

GBM =. Lx s. (1) 

Case II. — The Beam fixed at one end and the Load 

UNIFORMLY DISTRIBUTED OVER THE ENTIRE SPAN. 

The hending moment at any section is equal to the product 
of the load between the free end of tlie beam and the section 
by one half of its distance from the free end ; the Greatest 
Bending Moment, at the point of support, will be 



GBM:=^ L X 



o' 



(2) 



or one half what it is in Case I., the load, Z, and the span, .<?, 
being the same. 

Case III. — The Beam supported at the ends and the 
Load applied at the middle of the span. 

In order that eqiiilihriurn shall exist one half of the load 
must be supported by each point of support ; the load will 




bend the beam (Fig. 2) in the same manner that it would if 
the beam were Jixed in the middle of its span and loaded at 
each free end witli one half of the applied load, Z ; the he7id' 
ing moment at any section will be one half of the load 
multiplied by its distance from the nearest point of support ; 



CONCENTRATED PARALLEL FORCES. 

the Greatest Bending Moment, at the middle of the span, will 
be 

GBM=^^Xr,-^^, (3) 



or one fourth of that in Case I., smdone half that in Case II., 
the load and span being the same in each. 

Case IV. — The Beam supported at the ends and the 
Load, Z, uniformly distributed over the span. 

The Greatest Bending Moment occurs at the middle of the 
span, 



GB3f = 



Z X s 

8 



(4) 



Case V. — The Beam fixed at both ends and the Load 

APPLIED AT THE MIDDLE OF THE SPAN. 

In the preceding cases the bending moment of the applied 
load, Z, produces a strain of compression in either the top or 
the bottom of the beam and a tensile strain in the opposite 
side, but in this and the next case the bending moment pro- 




duces in the upper side, over each point of support, A A (Fig. 3), 
a tensile strain, and in the middle a compressive strain ; and 
in the lower side at each point of support, AA, a compressive 
strain, and in the middle, m, a tensile strain. [N^ow, in order 
that these directly opposite strains ma,y exist in the upper and 
lower sides of the beam at the same time, the strains at one 



6 STRENGTH OF BEAMS AND COLUMNS. 

section on each side of the middle section must be zero in 
intensity. At these points, rr^ the curvature of the beam 
changes ; they may be called the points of reverse curvature^ 
and in this case are located at one fourth of the span from 
each point of support, A A. 

The Bending Moments are equal and greatest at three 
sections ; at each point of support and at the middle of the 
span, theoretically its value is given by the following equa- 
tion: 

QBM = :^-^. (5) 

o 

Barlow and other experimenters state that this should be 

Z X s 



GBM=: 



6 



Case VI.— The Beam fixed at both ends and the Load 

DISTRIBUTED UNIFORMLY OVER THE SPAN. 

The sections at which the Greatest Bending Moment occurs 
are the same as in the preceding case ; but the points of re- 
verse curvature, rr, are at the distance 0.2113^ from each 
point of support, A A, Fig. 3. 

At the middle of the span, m, 

GBM^^^^' (6) 

24 

At the points of support, AA^ 

GBM = ^^' (7) 

6. Case in General. Theory has demonstrated and 
experiments fully confirm, except as previously noted, that the 
following relations exist between the Greatest Bending Mo- 
ments in beams, the span and the total applied load being 



CONCENTRATED PARALLEL FORCES. 7 

the same — that in a beam fixed at one end and loaded at the 
other being taken as our U7iit or standard of measure : 

n 

Beam fixed at one end and loaded at the other 1 

Beam '' " '' " " " uniformly ^ 

Beam supported at the ends and loaded at the middle \ 

Beam " '' " '' " " uniformly i 

Beam fixed at both ends and loaded at the middle ^ 

Beam " " " " " " uniformly yV 

From which we deduce the following formula, applicable to 
all of the preceding cases : 

GBM = Z X s X 71, (8) 

Placing for the factor, w, the values given in the preceding 
table of comparison, the formulas heretofore deduced for each 
case will be reproduced. 

These formulas for the Greatest Bending Moments are 
entirely independent of the material composing the beam and 
of its cross-section. 



8 



STKENGTH OF BEAMS AND COLUMNS. 



Section III. — Uniformly Varying Forces — Rectangular 

Areas. 

7. Notation. The principles governing tlie action of 
uniformly varying forces distributed over rectangular surfaces 
will be deduced, and then those applicable to surfaces bounded 
by curved lines will be considered. In order to determine the 
effect of such a force, we must determine the resultant, its 
point of application, its lever-arm, and from these the moment 
of the uniformly varying force. The following notation will 
be used, and it will have the same meaning whenever it appears 
in the following pages : 

C = the maximum compressive strain, in pounds, per square 

inch. 
T = the maximum tensile strain, in pounds, per square inch. 
J?^ z= the moment of the tensile strain in inch-pounds. 
R^ r= the " " " compressive strain in inch-pounds. 

(/^ =z the depth of the area covered by the compressive strain 

in inches. 
(I^ = the depth of the area covered by the tensile strain in 

inches. 
d = dc -\- d^, = the total depth of the area in inches. 
I = the total width of the area in inches. 

8. Resultant. The amount of direct strain is equal to 




the weight of a prismoidal wedge composed of such material 
that a prism an inch square in section and or T high will 



UNIFORMLY VARYING FORCES. 9 

give a pressure on its base equal to C or T as defined above ; 
the resultant will then be equal to the product of the depth 
rZc or d,^^ by the width J, by one halfoi the height C or T, which 
is the volume of the wedge, and it will pass through a point 
at f the depth d^ or d^ from the edge of the wedge A^ of 
which Fig. 4 is a section. 

. • . Resultant = — ^- or — -^' (9) 

Lever<i7'in = — ^ or -5-* V-'-'-'; 

o o 

By the Calculus : 

Let Xx — dx — AB, the depth of the area, 

X = any distance from A, 

Tx 

— = the height of the wedge at x, distance from A, 

it T 

hdx = a small area, or the differential, 

r^'-^Tx ^, Tbx'' 
Volume = I — oax = ~ — , 
Jq x"" 2^x 

. ' . Resultant = —^ > 

_,. ^ f^r Tx' Tbx^ 

The moment = J hdx = 7, — , 



. The moment = 



Tbd' 



8 
Dividing the moment by the resultant we obtain the lever arm = frfx. 

9. Problem I. Required the Moment of an uniformly 

VARYING compressive FOKCE OF MAXIMUM INTENSITY, C, WITH 
RESPECT TO AN AXIS, ^, IN THE BACK OF THE PRESSURE WEDGE. 

Let ABD (Fig. 5) represent a section through the pressure 
wedge. 

MM 



The resultant = 



c 



2 



-5 



10 



STRENGTH OF BEAMS AND COLUMNS. 



The lever-arm = ^d^ , 
• R = — — • 



(11) 




Problem II. Hequired the mortient of only a jportion of 
the force distributed over a depth EB = d^ and width Z*, — 
the force heing zero in intensity at A ; the axis heing at B 
(Fig. 5). 

d — d 

-^— — ^ • C = the intensity of the force at E^ 
cCq 

d 

— ' = the lever-arm of this force at E, considered 

to be constant over the distance EB^ 

M — - " ' }^ — *^^ intensity of the force at B less that 



d. 



2X E, 
- = its lever-arm. 



Having considered the force to be divided into two portions, 
the first, GE^ constant in intensity over the depth EB^ 
and the second increasing in intensity from zero at G to 

( 1 — -^-^ — ■- j (7 at i^, by adding the moments of these lyarts 

we obtain 



R. 



Qdr 



{Sd, - 2d,) a 



(12) 



UNIFORMLY VARYING FORCES. 



11 



By the Calculus : 

Let a-c = c?c = the depth AB (Fig. 5), 
' X = any distance from A, 

C = the intensity of the force at the depth x, 

Xc 

bdx = a small area of the base of the wedge, 

Jq X, \ 2xc dXc/ 

.-. Bc = -^, or Eq. 11. 

Integrating the above expression for Be, between the limits x = d^ and 
x = di, we obtain 

^'^'' {Mc - 2di) C, or Eq. 12. 



i?e = 



6c?c 



10, Problem III. Required the Moment, ^„ of an 

UNIFORMLY VARYING TENSILE FORCE, WITH RESPECT TO AN AXIS, 
(9, PARALLEL TO THE EDGE, A^ OF THE TENSION WEDGE, AND AT 
THE DISTANCE OA = d^ FROM IT. 

Fig. 6 




Let ABD (Fig. 6) represent a section through the tension 
wedge. 

hcl^. T . , 

— 7^ — = the resultant. 



or. 



f/e -}- i^h = its lever-arm, 



Jx^ 



hcL 



{U + d,) T. 



(13). 



12 STKENGTH OF BEAMS AND COLUMNS. 

Problem TV. Iiequired the moment of only a lym^t of the 
above force distributed over the depth EB = d^ and the 
width h^. 

Consider tlie force to be divided into two portions. The 
iirst, with the intensity at ^ constant over the distance EB ; 

the second increasing in intensity from zero 2ii G to -^ T 
atE. 

d^ (1 j-jT^ the intensity of the force at E, 



("-'!')=' 



its lever-arm, 



d 

-^ T = the intensity of the force at B, less that E, 

(d — — M = its lever-arm. 

By adding the moments we obtain 

B, = ^\^SdA i^d - d,) - d: (3rZ - 2«f,)]. (14) 

By the Calculus : 

Let a*! = ^T = the distance AB, 

X = any distance from A, 

icT 

— =: the intensity of the force at any distance x, 

(dc -\-x) = its lever-arm, 

bdx = the differential of the area of the base, 

. • . i?^ = -|-' {2d + dc) T, or Eci. 13. 

Integrating the above expression for i?T, between the limits x = dr — d-2 
and X = d/t, and placing h^ for b, we have 



UIS^IFORMLY VARYING FORCES. 



13 



6fZ, 



D 



i?^ = J^ r^drd^ {d - d,) - dj" {ddc - 2^2)."] T. 



This is not identical in form with Eq. 14, but gives the same numerical 
result. 



Section TV. — Uniformly Varying Forces — Circular 
Segment Areas. 

11. Resultant and Lever-arm. The resultant and 
the moment of an uniformly varying force, distributed ov^er 
circular segment areas, cannot be readily deduced by means 
of the elementary methods , used when the areas were rect- 
angular ; but this can be accomplished with the aid of the 
Calculus. 

12. Problem. Kequieed the moment of an uniformly 

VARYING FORCE DISTRIBUTED OVER A CIRCULAR SEGMENT, WITH 
RESPECT TO AN AXIS THAT IS A TANGENT TO THE BASE OF THE 
PRESSURE WEDGE AND PARALLEL TO ITS EDGE, WHICH IS A CHORD 
OF THE CIRCLE. 

Let, in Fig. 7, OBS represent the axis, MNB the base of 
the pressure wedge, and ABD a section through AB, 




Notation. — Adopting the notation defined in Art. 7 with 



the following in addition : 



14 STRENGTH OF BEAMS AND COLUMNS. 

Let r = the radius of the circle, 
d = d^ -\- dj: = the diameter, 
a?c = d^ = the versine of one half the arc MBN^ 
X := any distance from the axis OBS^ 

— C = the intensity of the pressure at any distance, 

a?, from the axis OBS, 

2 {2?'x — x^)^dx = the differential of the area of the base, 

Integrating, we obtain 

Substituting the following equalities : 

r versin. — = one half of the arc of the segment MBN 

Xq := CCq y 

reducing, we obtain 

^^'"^ 24dc\ ^^c^T l^(l\{dc-r) + r'^(30?'-14^e)] -^Comp.arc (12(?e- 15r)?'n(t5) 

from which the moment of any compressed wedge may be 
computed. The factor Comp. Arc is the arc MBN. 

13. Problem. Eequired the moment of an uniformly 

VARYING TENSILE FORCE, DISTRIBUTED OVER THE SEGMENT OF A 
CIRCLE, WITH RESPECT TO AN AXIS THAT IS A TANGENT TO THE 
CIRCLE, PARALLEL TO THE EDGE OF THE TENSION WEDGE, AND AT 
THE DISTANCE d^ FROM IT. 

Let, in Fig. 8, FOS represent the axis, MNB the base of 
the tension wedge, MN the edge, and OABD a section 
through OB. 



UNIFORMLY VARYING FORCES. 



15 




Let x^ — d^ — tlie distance AB, 

X = any distance from B toward A. 

^T — ^ . T — the intensity of the force at any distance x^ — x 

from A^ 
d — X — \i^ lever-arm. 

2 {^rx — x")^ dx — the differential of the area of the base of 
the wedge. 
For notation, refer to Arts. 7 and 12. 

E^ = f'^-^IiII^T{d -x)2 {2rx - x'fdx, 

«/ x^ 



r,E^ = ^T[{2rx.-x\)[^-^^ ^'"^ 



M~ "^ 24^/ 



^ r V 6 24 V J 



Substituting the following values : 



(2a^', - Q?\) =: 4/^/. 7' 



— 1 






Tversin. — =: one half the tension arc MBN. 



tC"J» ■ " rp • 

reducing, we obtain 



16 



STRENGTH OF BEAMS AND COLUMNS. 



R^ ^ 



24^/ L '^^c'^T [4<^\ (or — d^) + ISr' (r — d^)] 

+ Tension arc {12d^ — 9r)r^) 1 n g\ 

from wliicli the required moment may 
be computed. 

14. Problem. Kequired the re- 
sultant OR AMOUNT OF DIRECT FORCE OF 
AN UNIFORMLY VARYING FORCE DISTRIB- 
UTED OVER A CIRCULAR SEGMENT AREA. 

Let Fig. 9 represent the pressure 
wedge, MBN the base or circular seg- 
ment, and B the origin of co-ordinates. 
Let a?c = AB the versine of one half of the arc MBN^ 
X = any distance from the origin B^ 

k 
2 {^rx — x'') dx — the differential of tlie segment, 

T^= the resultant or volume of the pressure wedge, 

6^— the greatest height, DB^ 




'.^ca?. ^ Qy^ 2[^rx-x')dx, 



vbf. 



. • . \ =2C\ {^rx — x^) ( -^-! — \-{-rversin.-^[ - — - 1 • 

Substituting the equalities given in Art. 12 and reducing, we 
have 

-\- Segment arc {d^ — r) Zr • (17) 

The resultant of a tension force may be obtained from the 
above equation by placing T for C. 

The cubic contents of any cylindrical wedge may be com- 
puted from Eq. 17 by substituting for 6^, expressed in pounds, 
A, the greatest height of the wedge in inches. 



UNIFORMLY VARYING FORCES. 



17 



The area of tlie base of the pressure wedge or that of any 
segment of a circle may be computed from the following : 



fl 

Area = 2 \/cl^dj; (^c— ^) "h ^eg77ie7it arc Y.-f-, 

in which 

^4 = versine of one half the arc of the segment, 
r = the radius, 
d^ :=. the diameter, less d^. 



{\nA) 



Section N .— Uniformly Yarying Forces — Circular 

Arcs. 

15. Problem, Kequired the moment of an uniformly 

VARYING COMPRESSIVE FORCE DISTRIBUTED OVER THE ARC OF A 
CIRCLE, WITH RESPECT TO AN AXIS THAT IS A TANGENT TO THE 
CIRCLE AND PARALLEL TO THE CHORD JOINING THE EXTREMI- 
TIES OF THE ARC. 

Let, in Fig. 10, OBS represent the axis, MBN the arc, 
ABD a section through AB^ B the origin of co-ordinates, 
C the intensity of the force at B, and zero that at M and li. 




J) 



Fig. 1 



d, 



A 



For flotation refer to Arts. T and 12. 

Let Xc = dc — the versine AB oi one half the arc JJTBJV 
X = any variable distance from B or the axis. 



18 STRENGTH OF BEAMS AND C0LU3INS. 



C — the intensity of the force at x distance, 

Xq 

r = the radius of the circle, 
= twice the difierential of the arc, 



V2rx — X 

r> /"'^c 2/ 'ax ( a?c — x) p 



r 

Integrating, we obtain 

E, = 2Cr{2rx - x'jf^^^sjilM r - r) 

y ^Xq/ -J 

Substituting the vahies given in Art. 12, we have 
^„ = ^ [|/4^[2(3r - d^)] + Co?,>.p.arc{2d, - 3r)] - (18) 

16. Problem. Hequired the moment of an uniformly 

VARYING FORCE DISTRIBUTED OVER THE ARC OF A CIRCLE, WITH 
RESPECT TO AN AXIS THAT IS A TANGENT TO THE CIRCLE, PARAL- 



LEL TO THE CHORD CONNECTING THE EXTREMITIES OF THE ARC, 
AND AT THE DISTANCE cl^ FROM IT. 

Let, in Fig. 11, I^OS represent the axis, MBY the arc. 




rxgn 



Or 4-' 



rt 

T 

I 



B 



OABD a section tlirougli OB, 7'the intensity of the force at 
B, and zero at Jfand iT. 



UNIFORMLY VARYING FORCES. 19 

For notation refer to Arts. 7 and 12. 

Let a?T = cIt. = the versine of one half the arc MBN^ 
X = any distance from the origin B^ 



tJCr, iV 



T = the intensity of the force at any distance w, 



d — X := its lever-arm 
'^rdx 



— the differential of the arc, 



V2rx — xi 

2rdx 



i?, = f^^ _^^iL-_ (^^-ZJ^\ {d - X) 

JO y'^rx — x' V ^'x / 

Integrating, we obtain 
^T = -^ I Vd,d^ [2 {r + d,)] + Tension arc (2r7^- r) J, (19) 
from which the required moment may be computed. 

17. Problem. Hequiked the resultant or amount of 

DIRECT FORCE OF AN UNIFORMLY VARYING FORCE DISTRIBUTED 
OVER AN ARC OF A CIRCLE. 

Let, in Fig. 9, page 16, J/J5 3^ represent the arc of the circle, 
MNBD a wedge whose cylindrical surface is equal to the re- 
quired resultant, the force being C in intensity at B and zero 
at M and N. 

Let x^ = 6?c = the distance AB^ 

X — any distance from the origin of co-ordinates B^ 

J^ C = the intensity of the force at the point a?, 

Xq 

''Irdx 
./ o ^= the differential of the arc, 

Y := the resultant or volume of the pressure wedge, 

y _ rx^ X, — X ^^ "Irdx 
Jo X. 



^c V2rx — i 



X 



20 STRENGTH OF BEAMS AND COLUMNS. 

. • • ^=^i\j^ VdA+ ^V^^^^^ ^^^ K - r)!^, (20) 

from which the required resultant may be computed. 

The curved surface of a cylindrical wedge may be com- 
puted from the above formula by substituting for C^ expressed 
in pounds, A, the greatest height of the wedge in inches. 



CHAPTER II. 

RESISTANCE OF CROSS-SECTIONS TO RUPTURE. 

18. Moment of Resistance. The cross-section is 
the shape of the figure and the area that any material, such as 
a beam, would show, should it be cut into two pieces by a 
plane perpendicular to its length, and its resistance to rujpUire 
at this plane or section is the number of inch-pounds that its 
fibres Avill oifer to forces tendino; to cross-hrecik the beam or 
material of which it is a section : this is called the Moraent of 
Resistance of the cross-section. 

The Moment of Resistance varies in amount with the 
material and the shape of the cross-section, but it is entirely 
independent of the length of the beam and of the manner in 
which the load may be applied, in each case ; the same cross- 
section and material will offer the same number of inch-pounds 
of resistance when broken across. 

19. Neutral Line. AVhen a beam is broken across, or 
is acted upon by forces that bend or tend to break it, we know 




from observation that its fibres on the lower or convex side, 
AB (Fig. 12), are in a state of tension, and that those on the 



22 STRENGTH OF BEAMS AND COLUMNS. 

upper or concave side, CD^ are compressed ; but our knowl- 
edge obtained from observation is limited to what takes place 
on the surface of the beam. We can only know what takes 
])lace within such a beam by reasoning from analogy ; there is 
a tensile strain in the lower side of the beam, AB^ and just 
the reverse character of strain in the upper side, CD. In 
order that these two directly opposite strains may exist in the 
same beam at the same time, both strains must decrease from 
the surface toward a common point within tlie beam, where 
both strains become zero in intensity, and they may be classed 
and treated as uniformly varying forces. 

A line, 7^Z, for the longitudinal section of the beam, or a 
plane for the beam, is called the neutral line or line of no 
Htrain y its position in a beam having a cross-section of a 
given shape, at the instant of rapture, depends upon the 
material alone, or upon the ratio existing between the hreaJc- 
Ing compressive and tensile fibre strains. No line in this plane 
lias any of the properties of an axis that are usually assigned 
to it by writers on this subject. 

20. Bending Moment and Moment of Resist- 
ance. In order to obtain the relation existing between the 
Bending Moment of the applied load and the Moment of 
Resistance of the cross-section of the beam, conceive one 
half of the beam, ABCD (Fig. 12), to be removed and the 
bent-lever, og f ^\^. 13), to be substituted for it, and that the 
same cohesion to exist between the fibres of the bent-lever and 
those of the beam along the line, fg., that originally existed 
between the fibres of the two halves of the beam along the 
same line, the hent-lever^ however, preserving its distinctive 
(character of a hent-lever. The applied load., B, causes the 
l)ent-lever, og fi and the half of the beam, A gfD., to move 
downward in the direction of the lower arrow of the figure, 
and the end of the lever, 0, and that of the half of the beam, 
/), to revolve around f in the direction of the upper arrows, 



RESISTANCE OF CROSS-SECTIONS TO RUPTURE. 23 



/h 





T -^ 



FlQ 13 







the effect being identical with tliat produced by conceiving 
tlie fulcrum^ f^ to remain stationary and a moving force, 
Z -f- 2, to be applied to the end of the bent-lever, 0, and at 
the same time to be pressed in the direction of its length by 
a force, C^ equal in magnitude to the amount of direct com- 
pression along the line, fn^ above the neutral line^ nl. When 
the bent-lever is made to revolve around the fulcrum, y, it 
meets with an opposition of compression to its motion, de- 
creasing in intensity from ih^ fulcrum^ f^ to the neutral line^ 
n^ when it becomes zero in value ; at this point the opposition 
changes to a tensile resistance which increases in intensity 
from zero at n to its maximum, T^ at (j. 

The hent-lever and the original vertical section of the 
beam, fg^ are pressed closest together at \hQ fulcrum^ f -, from 
this point they continue to separate, by virtue of the ductility 
and compressibility of the material composing the beam, until 
rupture takes place, either in its upper or lower fibres ; the 
action of the bent-lever being identical with that of the half 
of the beam for wdiich it was substituted. 



21o Equilibrium. The Bending Moment of the applied 
load, Z, and the Moment of Resistance of the tensile and 



24 STRENGTH OF BEAMS AND COLUMNS. 

compressive fibre strains must be taken or computed witli 
reference to the fulcrum^ f^ and in order that equilibrium 
shall exist, the Bending Moment of the applied load, Z, must 
be equal to the Moment of Resistance, or the sum of the 
moments of the tensile and compressive fibre strains, and that 
the latter must be equal to each other in magnitude. 

22; Position of the Neutral Line. By deducing 
general formulas for the onoments of the tensile and compi^es- 
sive fibre strain in a cross-section of a given shape, placing 
them equal to each other and deducing from the equation so 
formed a general formula for either gn or fn, tlie position of 
the neutral line may be found in any section of the same 
shape by substituting in the formula its dimensions and any 
Jcnoivn values for the tensile and compressive fibre strains. 

In sections of an uniform shape, such as the rectangle and 
the circle, the depth of the neutral line below the compressed 
side of the beam may be obtained by multiplying the depth 
of the rectangle and the diameter of the circle by a deter- 
mined quantity that is constant for each shape and material ; 
this constant multiplier being its ]30sition when the depth of 
the rectangle and diameter of the circle is unity. 

But in irregular shaped sections, such as the T, Double T, 
Box, Rolled-eyebeain and other sliapes, in which the metal is 
not continuous from the neutral line to the top and bottom, a 
special solution must be made for eacli case to determine the 
position of the neutral line from which to compute the 
Moment of Resistance of the section. Before beams of this 
character are manufactured, an economical 2:)osition should be 
assumed and a sufficient area of metal placed above and 
below the neutral line to furnish the required Moment of 
Resistance. 

23. Neutral Line at the Transverse Elastic 
Limit. Our object in testing, to destruction, the strength of 
any piece of construction material is to obtain information 



RESISTANCE OF CROSS-SECTIONS TO RUPTURE. 25 

that will guide us to a correct knowledge of its use, when 
safety to life a.nd property is demanded, whether this destruc- 
tion be by means of extension or compression / as we apply 
the load in small instalments there are only two points in its 
intensity at which we can record the knowledge thus gained 
for future intelligent, comparative use when it has reached 
the elastic Umit, and when it is sufficiently intense to rupture 
or destroy the piece of material tested. When a beam is 
broken by a transverse load that has been applied in small 
instalments, we have two similar points, the elastic liirhit load 
and the rupturing load, and these are the only points at 
which our information can be used to compute the strength 
of similar material when used in structures. 

We know that in an ordinary beam without a load its com- 
pressive and tensile fibre strain is zero, and that the breaking 
transverse load produces the hreahing compressive and tensile 
strain in the fibres. Kow, does the transverse elastic limit 
load produce the elastic fibre strain limits % Is the neutral 
surface the same as that for rupture ? When the beam is un- 
loaded each plane of fibres is a neutral su'rface ; as the load 
is applied the compressive and tensile strain penetrates the 
beam from the top and bottom respectively ; theoretically we 
know they must meet at a common point within the beam at 
the instant of rupture^ and that this is fully 'sustained by ex- 
periments will be shown in the sequel. Our theory demands 
that when the same ratio exists between the tensile and com- 
pressive elastic fibre strain limits that does between the ulti- 
mate or breaking strains, in order that equilibrium shall exist, 
the neutral surfaces must be identical, but the theory does not 
require that the transverse elastic limit load shall j^roduce the 
elastic fibre strain limits ; we can only gain the desired informa- 
tion from discussing a numerical example. 

The mean compressive and tensile elastic fibre strain limits 
for good wrought-iron is C = 30000 pounds and T = 30000 
pounds per square inch. With these strains, from formulas de- 



26 STRENGTH OF BEAMS AND COLUMNS. 

duced in the sequel, the centre elastic limit transverse load of 
a bar of wrouglit-iron six inches square and ten feet span is 
44100 pounds. The centre elastic limit transverse load of a 
bar of wrought-iron, one inch square and one foot span, is 
2250 pounds, and that of the above beam from the well- 
known formula is, ' 

J , hd"^' 6 X 36 X 2250 ,_^^ . 

Load = —J— = - — — TTj = 48600 pounds. 

From this practical identity of results, as we have only used 
average values, and other special tests given in the sequel, we 
are authorized to conclude that the transverse elastic limit 
load produces the tensile and compressive fibre strain elastic 
limits. 

24. Movement of the Neutral Line with the De- 
flection. Having established the fact that the transverse 
elastic limit load produces the elastic fibre strain limits — and 
our theory requires that the elastic limit neutral line and the 
neutral line of rupture shall coincide only when their ratios 
are the same, but should they be unequal they must occupy 
diiferent positions — in the sequel it will be shown that where 
these ratios are unequal the elastic limit neutral line is situated 
hettoeen the neutral line of rupture and the bottom or ex- 
tended side of the beam, and that as the loading advanced 
from the elastic limit load to the rupturing load, the neutral 
surface must have moved upward or toward the compressed 
surface of the beam. 

From the above we conclude that as there was no change 
in the condition of the loading that could have reversed the 
direction in which the neutral line moved from its position at 
the elastic limit to that at rupture^ the neutral Ihie at the in- 
ception of the loading was at the bottom or extended side of 
the beam, and that as the loading progressed it 'moved up- 
ward or toward the compressed side of the beam— the tension 
area, to avoid rupture in its fibres, continues to encroach upon 



RESISTANCE OF CROSS-SECTIONS TO RUPTURE. 27 

the compressed area until the rupturing strain is produced in 
botli the top and the bottom of the beam. 

The neiitral line^ at the inception of the loading, being at 
the bottom or extended side of the beam, it can only be moved 
upward by reason of the deflection and equally with it. If, 
from the dimensions of the beam, it should not be able to 
deflect sufficiently to move the neutral line to the position 
required for equilibriuin between moments of resistance of 
the 'idtimate fibre strains, the true breaking strength will not 
be obtained for the beam. When this is the case the observed 
breaking load will be too large for wooden, wrought-iron, 
steel and tough cast-iron beams, and too small for the more 
fractious varieties of material ; for should the compressive 
strain reach its ultimate limit before the tensile strain an in- 
crease of the load will develop a crushing strain in excess of 
the true crushing intensity, as is frequently done in crushing 
short blocks; the beam will, however, continue to deflect 
under tliese increased loads, and will finally develop the full 
tensile strength, wdien the beam will be broken by a load 
much in excess of its true breaking load. 

From the above, the reason for the variation in the modulus 
of Txiptiire that is required in the '' common theory of 
flexure " is apparent, as the shorter beams in most series of 
experiments, especially of cast-iron, did not deflect sufficiently 
to break with the true breaking load, and, therefore, it re- 
quired a larger inodidus or empirical coefficient for the 
shorter beams than for the longer beams of the same series of 
tests. 

At the instant of deflection the hending ^noment of the 
applied load is held in equilibrium by a jpurely compressive 
resistance, distributed over the section as an uniformly vary- 
ing force, being zero in intensity at the bottom or extended 
side of the beam and greatest in intensity on the opposite side. 
This is a very important principle, as from it we shall, in tlie 
sequel, deduce the correct theory of the strength of columns. 



28 STRENGTH OF BEAMS AND COLUMNS. 

25. Neutral Lines of Rupture in Rectangular 
Sections. It maj not be out of place here to anticipate 
the resuks obtained in the sequel, by stating the positions of 
the neutral lines of rupture in a rectangular section when 
composed of the different kinds of material used in construc- 
tion. 

Let the section be six inches deep, q the ratio of the com- 
pressive to the tensile strains of rupture, or G ^ T. 

r depth of neutral line -^ 

^ ^ . r, K below the com- \ 

*^''*-"'°" ? = ^-5 1 pressed side of the \ 2-45 ins. 

I section J 

" q = 6. " " 3.00 " 

" q =z 4:. " " 3.24 " 

Steel ^ = 1.5 " " 4.29 " 

Wrought-iron q = 1. " " 4.68 " 

Beech, English....^ = 0.7Y5 " " 4.90 " 

" American q = 0.383 " " 5.36 " 

From this comparative statement we observe the order in 
which the neutral lines of rupture are arranged in rectangular 
sections ; the same order of arrangement exists in all other 
uniform sections. For different kinds of wood and cast-iron 
the neutral line of rupture lies between the extremes given in 
the above table of comparison. 

26. Relative Value of the Compressive and 
Tensile Strains. Experiments have fully shown that 
the compressive and tensile strains do not possess equal values 
as factors in determining the transverse load that a beam will 
bear, and that the influence of the tensile strain predomi- 
nates. 

The elastic limit or technical hreaMng load of a wrought- 
iron beam one inch square and twelve inches span, loaded in 
the middle, is 2000 pounds when 



HESISTAl^CE OF CROSS-SECTIOT^S TO RUPTURE. 29 

C = 30000, T = 30000 pounds, and the Moment of Ee- 
sistance = 6084 inch-pounds from our formulas. 

We will now endeavor to trace the effect that will be pro- 
duced upon the amount of the transverse breaking load from 
varying the exact and relative values of C and T from those 
in the above-described wrought-iron beam, which will be taken 
as our standard of comparison. Since the crushing and 
tensile strength of the material composing any beam must 
each sustain one half of the breaking load of the beam, the 
per cent of loss or gain in the transverse strength should be, 
approximately^ one half the sum of the per cents of the losses 
or gains in the values of C and J', but this will be found to 
be true for only the smaller ratios oi C -^ T that exist in 
materials of construction. 

The centre breaking load of a white pine beam one inch 
square and twelve inches span is 450 pounds when 

C = 5000, T = 10000 pounds, and the Moment of Ke- 
sistance = 1260 inch-pounds. In passing from the standard 
wrought-iron beam where C ^=^ T to the white pine beam 
where (7 = 0.5 T^ the following changes take place : 

Loss in the value of C 83.4 per cent. 

" T. m.Q '' 

Apparent loss to the transverse load 75.0 " 

Actual '' " " 77.5 " 

Loss to the Moment of Resistance 79.2 " 

results, practically, identical for this ratio. 

From Mr. Kirkaldy's experiments a bar of steel one inch 
square and twelve inches span will break with a centre load 
of 6400 pounds when 

C — 160000, T = 70000 pounds. Moment of Eesistance 
= 19200 inch-pounds. In passing from our standard wrought- 
iron beam to the steel beam, the following changes take 
place : 



30 STRENGTH OF BEAMS AND COLUMNS. 

Gain in the value of C .433.3 per cent. 

" " T 133.3 " 

Apparent gain to the transverse load 283.3 " 

Actual " " '^ '' 220.0 " 

Gain to the Moment of Resistance 215.0 " 

Mr. Hodgkinson found the centre breaking load of a certain 
cast-iron beam, one inch square and twelve inches span, to be 
2000 pounds when 

C = 115000, T = 14200 pounds, and the Moment of Ee- 
sistance = 6600 inch-pounds. In passing from our standard 
wrought-iron to the cast-iron beam of the same size, the 
following changes take place : 

Gain in the value of C -|- 283.4 per cent. 

Loss " " " '' T - 52.6 " " 

Apparent gain to the transverse load .... 115.4 " " 

Actual " '^ " " 0.0 " " 

Gain to the Moment of Resistance 8.4 " " 

In this experiment it required 283.4 per cent gain in the 
value of C to offset a loss of 52.6 per cent in the value of T^ 
or that the compressive strength does not sustain its proper 
proportion of tlie load. 

This great discrepancy between the legitimate theoretical 
deductions and the results obtained from experiments cannot 
be reconciled on the hypothesis that the forces are in equi- 
librium with respect to an axis that lies within the beam, 
the moment in each case, for rectangular sections, being the 
resultants of the tensile and compressive forces, multiplied by 
two thirds of the depth of their respective areas, showing that 
the compressive strain works under no disadvantages ; but on 
our theory this discrepancy is fully accounted for. The lever- 
arm of the crusJung resultant is one third the depth of the 



RESISTANCE OF CKOSS-SECTIONS TO RUPTURE 31 

compressed area, while that of the tensile resultant is only 
one third less than the total depth of rectangular beams. 

27. Suiiimary of the Theory. The theory herein 
advanced to explain the relation that exists between the Bend- 
ing Moment of the applied load and the Moment of Resist- 
ance of the material composing the beam, may be expressed by 
the following hypotheses : 

1st. Tlie fibres of the beam on its convex side are extended 
and those on the concave side are compressed in the direction 
of the length of the beam, and there are no strains but those 
of extension and compression. 

2d. There is a layer or plane of fibres between the extended 
and compressed sides of the beam that is neither extended nor 
compressed, w^hich is called the neutral surface or neutral line 
for any line in this plane. 

3d. The strains of compression and extension in the fibres of 
the beam are, in intensity, directly proportional to their dis- 
tance from the neutral surface. 

4th. The axis or origin of moments for the tensile and com- 
pressive resistance of the fibres of any section at right angles 
to length of the beam, is a line of the section at its intersection 
with the top or compressed side of the beam. 

5th. The fibres of a beam wall be ruptured by either the 
tensile or compressive strains in its concave and convex sur- 
faces, whenever they reach in intensity those found by experi- 
ment to be the direct breaking tensile and compressive fibre 
strains for the material composing the beam. 

6th. The Bending Moment of the load at any section is 
equal to the sum of the moments of resistance to compression 
and extension of the fibres, or to the Moment of Resistance of 
the secti/)n of the beam. 

7th. The sum of the moments of resistance of the fibres to 
compression is equal to the sum of the moments of resistance 
of the fibres to extension. 



32 STRENGTH OF BEAMS AND COLUMNS. 

8tli. The algebraic sum of tlie direct forces of compression 
and extension can never become zero. 

9th. The Moment of Resistance of the section is equal to 
the sum of the moments of resistance to the compression and 
extension of its fibres. 

10th. The transverse elastic limit load produces the tensile 
and compressive fibre strain elastic limits. 



CHAPTER III. 

TRANSVERSE STRENGTH. 
Section I. — General Conditions. 

28. Coefficients or Modvili of Streng^th are quanti- 
ties expressing the intensity of the strain under which a piece 
of a given material gives way when strained in a given 
manner, such intensity being expressed in units of weight for 
each unit of sectional area of the material over wliich the 
strain is distributed. The unit of weight ordinarily employed 
in expressing the strength of materials is the nuinber ofjjonnds 
a/voirduj)ois on the square inch. 

Coefficients of Strength are of as many different kinds as 
there are different ways of breaking a piece of material. 

Coefficients of Tensile Strength or Tenacity is the strain 
necessary to rupture or pull apart a prismatic bar of any given 
material whose section is one square inch, w^hen acting in the 
direction of the length of the bar. This strain is the T oi 
our formulas. 

Coefficient of Crushing Strength or Compression is the 
pressure required to crush a prism of a given material whose 
section is one square inch, and whose length does not exceed 
from one to Jive times its diameter, in order that there may be 
no tendency to give way by bending sideways. This pressure 
is the C of our formulas. 

29. Elasticity of Materials. It is found by experi- 
ment that if the load necessary to produce a strain and fracture 
of a given kind is applied in small instalments, that before 
the load becomes sufficiently intense to produce rupture, it will 



34 strength: of beams at^d colum:n^s. 

cause a change to take place in the form of the material, and 
if the load is removed before this intensity of the fibre strain 
passes certain limits, the material possesses the power of re- 
turning to its original form. This is called its elasticity. 

30. Elastic Limits. When the material possesses the 
power of recovering its exact original form without " set" on 
the removal of a load of a given intensity, the greatest load 
under which it will do this is called the limit of perfect elas- 
ticity. 

The limit of elasticity as ordinarily defined and used by 
experimenters is that point or intensity of strain where equal 
instalments or increments of the applied load cease to pro- 
duce equal changes of form, or where the change in form in- 
creases more rapidly than the load. 

31. The Elastic Limit of Beams may be deter- 
mined by applying small equal parts of the load and noting 
the increase in deflection after each increase of the load, al- 
lowing sufficient time for each increase of the load to pro- 
duce its full effect. When it is found that the deflections 
increase more rapidly than the load, its elastic limit has been 
reached and passed. The relation between the elastic limit 
load of the beam and the elastic limit of the tensile and com- 
pressive fibre strains of the material composing the beam will 
be shown in the sequel, or that the elastic limit load of the 
beam produces the elastic limit strain for the fibres. 

32. Working Load and Factor of Safety. 

The greatest load that any piece of material, used in a struc- 
ture, is expected to bear is called the worMng load. 

The hreaking load to be provided for in designing a piece 
of material to be used in a structure is made greater than the 
worMng load in a certain ratio that is detennined from ex- 
perience, in order to provide for unforeseen defects in the 
material and a possible increase in the magnitude of the ex- 
pected working load. 



TRANSVERSE STRENGTH. 85 

The factor of safety is the ratio or quotient obtained by 
dividing the hreahing load by the worthing load required. 

33. General Formula. In our first chapter we 
deduced rules or formulas, from which can be computed the 
Greatest Bending Moment that a load applied to a beam in a 
given manner will produce without reference to the shape of 
its cross-section, or to the material composing the beam. 

In our second chapter i)rinciijles are deduced from which 
can be computed the Greatest Moment of Resistance cross- 
sections of the various shapes and material will exert at the 
instant of rupture, without reference to the length of the beam 
or to the manner in which the load may be applied. 

To avoid repetition, the formulas for the Moments of Re- 
sistance are deduced in this chapter. Our knowledge of the 
transverse strength of beams will now be complete if we com- 
pute and make the Greatest Moment of Resistance of the 
cross-section of the beam equal to the Greatest Bending 
Moment of the applied load. 

Let i? = the Greatest Moment of Resistance of the l)eam, 
L = the total applied load in pounds, 
s = the span, the distance between the supports in 

inches, 
n = the factor defined in Art. 6, page 7, 
M = the Greatest Bending Moment of the applied 
load; 



(21) 
from which the breaking load of the beam may be computed. 

34. Relative Transverse Strength of a Beam. 

Referring to the values of the factor, n, of Eq. 21, as given 



Eq. 


8, 


page 


7, 


we have 
. Z = 


nLs 
R 

ns 



36 STRENGTH OF BEAMS AND COLUMNS. 

in Art. 0, page 7, it is found in each ease to be either 
unity or a fraction^ and when these values are introduced for 
n^ in Eq. 21, it is equivalent to multiplying the numerator, R^ 
by the denominator of the fraction, n ; hence if we make 

1 

— = m, 
n 

Eq. 21 will become 

Z = — . (22) 

s 

Computing and tabulating the values of m from those of /i, 
Art. 6, we obtain the following relation between the breaking 
loads of a beam, the span, material and cross-section remain- 
ing the same or constant ; the breaking load of a beam, fixed 
at one end and loaded at the other, being the unit or standard 

of strength. 

m 

Beam fixed at one end and loaded at the other 1 

Beam '' " " " uniformly... 2 

Beam supported at its ends and loaded at its middle 4 

Beam " " " " " uniformly 8 

Beam fixed at both ends and loaded at its middle 8 

Beam " " " " " uniformly 12 

Eq. 22 is the general formula from which will be de- 
duced the transverse strength of beams of all sections by 
giving to the factor, R, its proper value. 



TRANSVERSE STRENGTH. 



37 



Section II. — Transverse Strength — Rectangular Sections. 

35. Moment of Resistance. In Fig. 14, let ^^i>X 
represent the section, nl the neutral line, AnlX the compressed 
area, BnlD the extended area, AnBMnNA any section through 




the strain wedges at right angles to the neutral line, and AX 

Adopting the notation heretofore used in 



the axis or origin. 



Art. 



7, we have 



C = the greatest intensity, per square inch, in peunds of 
the compressive strain, 

T = the greatest intensity, per square inch, in pounds of 
the tensile strain, 

H^ =: the moment of the compressive strain in inch-pounds, 

R^ = the moment of the tensile strain in inch-pounds, 

R =^ R^-\- Rj, = the Moment of Resistance of the section in 

inch pounds, 
r/c = the depth of the compressed area. An, in inches, 
d.^, =: the depth of the extended area, Rn, in inches, 
d = 6?c + ^^4 = tl^e depth of the section AR in inches, 
h = the width of the section RR in inches, 
q =^ the quotient arising from dividing O by T, 

L = the total applied load in pounds, 
s = the span, the distance between the supports in inches, 

m = the factor, defined in Art. 34, page 35. 



38 STRENGTH OF BEAMS AND COLUMNS. 

From Eq. 11, page 10, we have, for the moment of com- 
pressive resistance, 

i?c = *^, (23) 

and for the moment of the tensile resistance from Eq. 13, 

-R^^^-pi^d+cQT. . (24) 

36. The Neutral Line, From the 7th hypothesis, Art. 
23, page 31, we have, from equating the second members of 
Eqs. 23 and 24, 

6 6 ^ 1 c; J 

from which 



and 






c\ = ^(-0>5 + 1/2^+2.25) ^26) 

The position of the neutral line in any rectangular sec- 
tion may be computed from either of these formulas ; it is 
independent of the absolute intensity of the maximum tensile 
and compressive iibre strains, but depends upon their ratio 
C-^ T— q for its position. The application of these formulas 
is illustrated in Examples 5, 6 and 7 of the sequel. 

37, Transverse Strength, General formulas for the 
transverse strength may be obtained by substituting for 7? the 
Moment of Eesistance in Eq. 22, page 36, its values 2i?c == 27?^ 
from Eqs. 23 and 24. 

• '^ 3^-' ^^'^ 

. • . Z = '"^^L {2d + d.) T, (28) 

OS 



TRANSVERSE STRENGTH. 39 

From eitlier of these equations, either the breaking load or 
the elastic limit load may be computed by giving to C and T 
the breaking or elastic limit values of the material composing 
the beam. The application of Eq. 27 is illustrated in Exam- 
ples 7, 26, 27, 28, 36, 37, 38, 39 and 40, and Eq. 28 in Ex- 
amples 5, 6, 7, 8, 25 and 30 of the sequel. 

38. To Design a Beam. In designing a beam of a 
rectangular section that shall break with a given total applied 
load, distributed over the span of the beam in a given manner, 
the span of the beam will be determined from the position in 
which it is to be used, and the depth, cZ, in inches will be as- 
sumed / the criLsliimj and tensile strength of the material 
composing the beam must be hiown / it will then only be nec- 
essary to determine the width^ Z>, in order that the beam shall 
break with the required load. 

From Eq. 25 deduce the position of the neutral line^ which 
is independent of the width, 5, then from Eq. 27 deduce the 
value of h, the width 

•••^=^„ (29) 

which gives the required width, since all of the factors in the 
second member of this equation are known quantities. 

Should the assumed (iQ^t\i and the computed ^Y\di\\ not give 
an economical section for the beam, a second depth must be as- 
sumed from this information and a second width computed ; 
this process should be repeated until a satisfactory result is 
secured. 

39. To Compute the Compressive and Tensile 
Strains. 

Pkoblem I. — The position of the neutral line and the 
crushing strain of the material of a rectangidar section may 
he computed from the knotvn transverse hreaking load and 
tensile strength. 



40 STRENGTH OF BEAMS AND COLUMNS. 

From Eq. 28, page 38, deduce tlie value of r/„ 

and 

which gives the required position of the neutral line, illus- 
trated in Examples 1, 2, 21, 22 and 35. 

The crusJiing strength can be computed by deducing its 
value, 6', from Eq. 27, 

mbd: ^ ^ 

The application is illustrated in Exaniples 1, 2 and 21. 

Problem II. — The position of the neutral line and the 
tensile strength of the material of a recto/ngular section may 
he com^puted from the known transverse hreaking load of the 
heam and the crushing strength of the matei'ial. 

From Eq. 27 we have 

a, ^ Z-^. (32) 

mo (J 

for the position of the neutral line, which is illustrated in Ex- 
ample 4. 

From Eq. 28 we have for the required tensile strength 

rp O l^S /"QQX 

"" mhd^ {M -f d)' ^ ^ 

illustrated in Example 4. 



TRANSVERSE STRENGTH. 



41 



Section 111. — Transverse Strength — Ilodghlnson Section. 

40. Moment of Resistance. In Fig. 15 let ABDX 
represent the section, AOX tlie axis or origin of moments. 



Ji 





y^ 




I 


1 

-1- 


, 6. 

t 






^C 







B 



D 




nl the neutral line, the area above it being convpressed and 
that below it extended. 

OnEMnNO^ a section through the strain wedges on the 
line OE. 

d = the depth of the web and section, OE, in inches, 

d^ =z the " " npper flanges in inches, 

d^ — the " " lower a u 

h = the width of the web in inches, 

h^ = the sum of the widths of the upper flanges in inches, 

or AX - 5, 
^2 = the sum of the widths of the lower flanges in inches, 

or BI) - h. 
For other notation refer to Art. 35. 

From Eqs. 11 and 12, page 10, we have for the moments of 
com^pressive resistance, 



Web, E, = 



6 ' 



7 7 2 

Upper flanges, E^ = -^-j- {^d — 2d^) C. 



42 STRENGTH OF BEAMS AND COLUMNS. 

Adding, we obtain for the section, 

and for the ?no?nents of tensile resistance from Eqs. 13 and 
14, page 11, 

Web, E, = ^-^ {2d + d,) r, 

Lower flanges, ^. = ^ \j>dA {U - d,) — d,' (Sd - 2^7,) 1 T. 

Adding the above equations we obtain for the moinent of 
tensile resistance of the section, 

7?,= VM^\U-\-d^ ■^llM.dJ^U-d^-dX^d-M,)]]^ (35) 

and the Moment of Resistance of the Hodgkinson section is 
2^e = 2/4 from Eqs. 34 and 35. 

41. The Nevitral Line. By equating the vakies of 
R^ and 7?T given l)y Eqs. 34 and 35, and deducing from the 
equation so formed the vahie of d^^ the position of the neutral 
line is determined, but as this will involve the solution of a 
biquadratic equation, the general solution will be too complex 
for ordinary use. 

The following approximate formula obtained by neglecting 
certain quantities that do not materially affect the result in 
ordinary cases, will give its position sufficiently near for all 
practical purposes when the beam is made of cast-iron : 



, ^ <n + V\-2bddJ,, {q + 1) +<.Vf (4y + 5) (oas 

Its application is illustrated in Examples 8, 0, 10 and 29. 

The position of the neutral line in a Hodgkinson or in 
any single or double flanged beam may be computed when 
the transverse strength of the flanged beam and either the 



TRANSVEESE STRENGTH. 43 

compressive or tensile strength of the material have been deter- 
mined experimentally. 

Problem I. — Required the position of the neutral line in 
any flanged heam when the transverse^ ultimate^ or elastic 
load and the corresponding comjpressive strength of the 
material have heen obtained from encperiinents. 

For any given flanged beam with its transverse and com- 
pressive strength determined experimentally, all of the terms of 
Eq. 37, page 44, except d^^ become known quantities ; by 
giving the letters their numerical values for any beam, the 
formula can always be reduced to the following general form : 

d' ± Sjjd^ ± 2k = 0, 

in which 3j? and 2k are numerical quantities ; then by Car- 
dan's Kule for the solution of cubic equations of this form, 

(^c=- yk+ |/>F+^' + V^- - l//^' + p% (A 36) 

p and k must be made equal to one third and one half of 
the numerical quantities, 3/> and 2k, respectively, and their 
algebraic sign must be that of the term in which tliey appear 
in the reduced equation. 

From the above equation one value of d^ will be determined, 
and with it, Eq. (A 36) may be reduced to an equation of the 
second degree by dividing the cubic equation by d^ plus or 
minus the numerical value deduced above, giving the numer- 
ical value in the division the reverse sign to that computed. 
From this resulting equation of the second degree, the other 
two values of d^ may be computed ; the value that represents 
the position of the neutral Hue may be generally determined 
from inspection. 

Problem II. — Required the position of the neutral line in 
any flanged heam when the transverse load and tensile 
strength of the material have heen obtained from experiments. 

In Eq. 38, wdiich gives the relation between the Moment of 



44 STRENGTH OF BEAMS AND COLUMNS. 

the applied load and the Moment of Resistance of the section, 
all of the terms become known quantities except d^ ; substitut- 
ing for the letters their values in any given case, it will 
always reduce to the following general form : 

d^ — ad^ ± ind^ ± ^^ = 0, 

in which a^ in^ and n are numerical quantities. By substitut- 
ing for 6?T a new unknown quantity, 

(Ij. — X — |— — , 

the equation will reduce to the following general form : 

x' ± Zpx ± 2A: = 0, (B 36) 

which can be solved by Cardan's Rule, as in the preceding 
Problem. 

42. Transverse Strength, To obtain a general 
formula from which the transverse strength of a Hodgkinson 
beam may be computed, place for R^ the Moment of Resist- 
ance in Eq. 22, page 36, its values, 2^^ = 2^^? from Eqs. 34 
and 35, and we have 



and 



L = |j^.[w; + h,d; {3d, - 2fZ,)], (37) 



Z=~\'hd^'' {2d-\-d,)+b,[3d,d^{2d-d,)-d,' {Zd-M,)]]{SS) 
oaT^sL. J 

From either of these formulas, either the breaking or elastic 
limit load may be computed. Their application is illustrated 
in Examples 8, 9, 10, 22, 29, 30 and 31. 

43. To Design a Hodgkinson Beam. An econom- 
ical position for the neutral line must be assumed and a suffi- 
cient area of the metal arranged above and below it, to furnish 
the required cotnjpressive and tensile resistance / the depth of 



TRANSVERSE STRENGTH. 45 

tlie beam, d^ and the thickness of the web, J, should also be 
assumed. 

Let L = the required breaking load in pounds, 
s = the span in inches, 
TYh =: the factor defined in Art. 35. 

Step I. — Assume the thickness of the web, J, and its 
depth, d^ which is also the depth of the beam, and an economi- 
cal position for tlie neutral line. The compressed area above 
the neutral line must furnish one half and the extended area 
below it the other half of the required strength of the beam. 

For the compressed area : 

Step II, — Place the value of the compressive resistance 
given by Eq. 11, page 10, for 7? in Eq. 22, page 36, and we 
shall have for the load, Z„ sustained by the compressed area of 
the web^ 

A - —^^' (39) 

Step III. — For the top flanges deduct the load, Zj, ob- 
tained in Step XL, from one half of the required breaking 
load, Z, of the beam ; the balance is the breaking load, Z^, for 
the top flanges, or, 

A - f - A- m 

In order to design an area of section that will sustain this 
load, we must assume a convenient depth, c?,, for the top 
flanges and obtain the sum of their widths, 5,, by placing for 
R^ in Eq. 22, page 36, its value for this case, from Eq. 12, page 
10, and we will have 

J _ l,<i: (3.7, - 2^,) C 



from which we have 



7, — ^dcL s ,.2\ 

' d,\Zd,-M)mC' ^ ^ 



46 STRENGTH OF BEAMS AND COLUMNS. 

One half of this computed width, J„ must be arranged on 
each side of the web. 

For the extended area : 

Step IV, — Substitute for 7?, in Eq. 22, page 36, its value 
from Eq. 13, page 11, and we will have for the load, Z3, sus- 
tained by the extended area of the web, 

Z3 = M,(2ri + <^e)?^. (4-3) 

Step V» — For the hottom flanges deduct the load, Z., 
found by Step lY., from one half of the required breaking 
load, Z, of the beam ; the remainder will be the load, Z^, that 
must be sustained by the bottom flanges, or 

L. = r^- L. (44) 

In order to design an area for the hottom flanges that will 
sustain this load Z^, we must assume a convenient depth, fZ^, 
for the bottom flanges, and obtain the sum of their widths, J^, 
by substituting for R^ in Eq. 22, its value for this case from 
Eq. 14, page 12, and we will have 

L, = \ [3.7/7, {U - <) - d: (3^ - 2^0] '^, (^'^) 

from which we deduce 

7, — hd^L^ , ^. 

' \?>d^d^ (2^7 - d,) - d: (Sd - 2d,)] mT ^ ' 

One half the width computed from this formula must l>e 
arranged on each side of the web. 

Step VI. — Should the computed dimensions from those 
assumed produce a badly designed section, new dimensions 
must be assumed from the knowledge thus obtained, and a 
second computation made as in the first instance. 



TRANSVEESE STRENGTH. 



47 



Section IV. — Transverse Strength — Double T and Hollow 
Rectangle or Box Section. 

44. Moinent of Resistance. Let Fig. 16 represent 
the sections of tlie Double T and Box Sections respectively ; the 
same letters in the text apply to both sections. 



A 







^ 




. h 


I 


I- 


^ 

^ 




^ 



T 




At ->E B. 





aj 




... 






h. 






J^ 





D 



Let ABDX represent the sections, OnEMnNO a section 
through the strain wedge on the line OE^ A OX the axes, and 
nl the neutral lines. 

Notation for the Double T : 

1) = the width of the web in inches, 
l)^ = 1).^ = the sum of the widths of the flanges of tlie double 
T in inches, or ^^ — h, 
d^ = the depth of the top flanges of the double T, 
d, = the " " " bottom " " '' 

Notation for the Box Section : 

1) =z the sum of the widths of the sides of the box in 
inches, 
h^ =z h^ =z the inside width of the box, 

rZj =r the depth of the top of the box, 

d„ = the " " " bottom of the box. 



Giving the letters the above definition the moments oi tensile 



48 STRENGTH OF BEAMS AND COLUMNS. 

and compressive resistance may be obtained from the following 
formulas : 

B, from Eq. 34, (47) 

E^ '' " 35. (48) 

45. Neutral Line. The position of the neutral line in 
the Double T and Box Sections, when constructed of cast-iron, 
may be computed approximately from the formula given for 
the neutral line in the Hodgkinson beam, or 

< from Eq. 36. (49) 

When beams of these sections are made of wrought-iron and 
steel, the neutral line is so close to the bottom of the section 
that its position cannot be determined from an approximate 
formula. An exact solution of a long equation of the fourth 
degree must be made to determine its position. 

46. Transverse Strength. The breaking strength, 
Z, may be computed from the formula given for the trans- 
verse strength of the Hodgkinson beam, by noting the defini- 
tion of the letters given in Arts. 35 and 44. 

L from Eq. 37, (50) 

L " " 38. (51) 

47. To Design a Double T and Box Section, 
Step I. Assume the depths <:Z, of the Box or Double T and 

the position of the neutral line, also the width, h, of the web 
of the Double T, or the sum of the equal widths of the sides of 
the box. The compressed area above the assumed position of 
the neutral line must sustain one half and the extended area 
below it the other half of the Bending Moment of the applied 
load, Z, that the beam is required- to break with. 

For the compressed area : 

Step II. — The moment of compressive resistance that the 
sides of the Box or the web of the Double T will oifer to the 



TEANSVERSE STRENGTH. 49 

Bending Moment of the applied load from Eq. 11, page 10, 
being substituted for B, in Eq. 22, page 36, will give for its 
proportion of the load 

Z. = "^^. (52) 

Step III. — For the to]) flanges of the Double T or the 
tojp of the Box Section deduct the load, Zj, found by Step II. 
from one half of the total ap]3lied load, Z ; the remainder 
will be the breaking load, Z^, for the top flanges or the top 
of the box, as the case may require, or 

A = I - L,. (53) 

To design an area of section that will resist this load, Z^, 
assume a width, ^j, for the sum of the widths of the top flanges 
or the top of the Box, and compute therefrom their depth, d„ 
by substituting for the unoment of C07npressive resistance^ R^ 
in Eq. 22, page 36, its value in this case from Eq. 12, page 10, 
and we will have 

A = 14: {^<h - m "ff^, (54) 

from which deducing the value of d, by making d^"" =z d^\ 
which may be done without appreciably affecting the result 
obtained, we have 



d^ =. |/. ^^'■^"'^ 



(55) 



b, (34-2) mC 
from which the required depth may be computed. 

EOK THE EXTENDED AREA : ' 

Step IV. — The moment of the tensile resistance that the 
web of the Double T or the sides of the Box Section will 
offer to the breaking moment of the applied load, Z, will be 



50 STRENGTH OF BEAMS AND COLUMNS. 

obtained from Eq. 13, page 11, wliicli being placed for i? in 
Eq. 22, page 36, gives for its proportion of the load, Zg, 

Z3 =. K {M + rZe) ~, (56) 

Step V. — The hottom flanges of the Double T or the 
hottom of the Box Section must sustain as its proportion of the 
load, L^^ 

A = I - A- (57) 

In order to design an area that will sustain this load, Z^, 
assume a width, h^ = Jj, the width assumed in Step III., and 
compute the corresponding depth, d^^ by placing for i?, in 
Eq. 22, the value of the moment of the tensile resistance for 
this case, from Eq. li, page 12, and we have 

Z, = [3^^. (2rZ - d,) - d: {3d - M,)~j 1^, (58) 

from which, deducing the value of d^, by making cZ/ = d^^ 
we have 



'~ 2b^mTCMr + M-\-'2) ' ^^ 

from which the required depth may be computed. 

Step VI. — Should the section obtained by this process 
not be satisfactory new dimensions must be assumed to remedy 
the defects of form, and the process repeated. 

Cast-Iron Double T and Box Sections : 

For the Box or Hollow rectangidar beam the assumed 
widths, 1)^ = ^25 must be placed between the sides of the box, 
that having the depth, 6?„ from Eq, 55, must be placed at the 
top, and <^2, computed from Eq. 59, at the bottom of the Box 
beam. 



TRANSVERSE STRENGTH. 51 

For the Double T beam the assumed width, h^ = h^, must 
be placed in equal projecting flanges on each side of the web 
at the top and bottom, that having the depth d^ at the top 
and d^ at the bottom of the beam. 

Wrought-Iron Double T and Box Section. 

For the Box or Hollow rectangular beam made of riveted 
plates a part of the assumed width, h^ = l>^^ must be placed be- 
tween the sides of the Box and the balance in two equal project- 
ing flanges on the outside of the Box at the top and bottom ; the 
total width of the rolled plate that forms tli« top and bottom 
of the Box is equal to (J -f- Jj), and the depth of the plates 
that form its sides is equal to cZ — (d^ -\- d^ as found by Steps 
III. and y. 

For the Double T : 

Riveted plate sections. The width of the top and bottom 
plates is equal to {b -\- b^), and the depth of the web plate is 
d — {d, + d,). 

Rolled Eyebeams are arranged like that for a cast-iron 
Double T beam. 



52 



STRENGTH OF BEAMS AND COLUMNS. 



Section Y. — Transverse Strength. 

The Inverted T, Double Inverted T and |_J Sections. 

48. Moment of Resistance. Let Figs. 17, 18 and 
19 respectively represent the three sections, nl the neutral 
lines, and AXt]iQ agues or origin of moments of resistance for 
each section. 



^_X 



A 



X 



Tx^u 


\ 


'1^ 

.1 

I, 

. • - -1 ^ 


^ 




■^ J. 


a 




Nl/ 



^l. 



^ 



c 



Fi^J 8 



t 



J\ 



X 



.1 "M. 



V 



\ 



Fig. 19 



D 



M 



N 



In addition to the notation given in Art. 35, observe the 
following for nse in the formulas given in this Section : 

The Inverted T, Fig. 17. 

h = the width of the web in inches, 

h.^ — the sum of the widths of tlie flanges-, or BD — J, 

d^ = the depth of the flanges in inches. 

The Donhle Inverted T, Fig. 18. 

h z=z tlie sum of the widths of the webs in inches. 



h 

d„ 



flanges, or MJV — h, 



the depth of the flanges in inches. 

The LI Section, Fig. 19. 

h = tlie sum of the widths of the webs in inches, 
h^ — the widtli of the bottom in inches, or OP — h, 
d^ = the depth of the bottom in inches. 

Noting tlie above definitions, the moment of compressive 
resistance for each section will be from Eq. 11, page 10, 



rCo = — 7. — , 



(60) 



TRAI^SVERSE STRENGTH. 53 

and for the mom£7it of tensile resistance tlie formula given 
for the Hodgkinson beam will apply, or 

R^ from Eq. 35, page 42. (61) 

The Moment of Resistance of the section will be 

B^ jR,-{-B., = 27?, = 27?„ (62) 

from the above equations. 

49. The Neutral Line. In cast-iron beams of these 
sections the neutral line will be given approximately by the 
following formula : 

. _ dh + VmdbA (g + 1) + f^'d' (4g + 5) 

50. Transverse Strength. Place for E in Eq. 22 

the values for the Moment of Resistance of the sections, and 
we will have 

L = -3^-. (64) 

and 

L from Eq. 38, page 44. (65) 

From either of these formulas the transverse breaking load 
may be computed. 

51. To Design an Inverted T, Double In- 
verted T and i_| Sections. 

Stei> I. — Assume a value for the depth, f7, and the width 
or sum of the widths of the webs, J, and an economical posi- 
tion for the neutral line^ and from these compute the othei- 
dimensions. 

Step II. — The compressed area above the neutral line 
must have a moment of resistance equal to one half the Bend- 
ing Moment of the applied load, Z, or 

hd^^ G Ls rr^c^. 



54 STREI^GTH OF BEAMS AND COLUMNS. 

If our assumed values for h and d^ do not satisfy this equa- 
tion, other values must be assumed until the two members 
give identical numerical quantities. 

For the extended area : 

Step III. — The load, Z, sustained by the extended area of 
the web or webs, ^, may be obtained by substituting their mo- 
7nents of resistance for H in Eq. 22, page 36, its value from 
Eq. 13, page 11, 

L, = 'bd,{U + dy'^. (67) 

Step IV, — For the flanges and hottom deduct the load Z, 
found by Step III., from one half of the applied load Z, the 
remainder, L^^ is the load that must be sustained by the flanges 
and bottom. 

A = ^ - A- (68) 

In order to design an area that will sustain this portion Z, 
of the applied Z, assume a convenient depth, d^^ and from this 
compute the width h^^ by substituting for B in Eq. 22, page 
36, its value, the moment of tensile resistance for this case 
from Eq. 14, page 12, 

Z, = \UA {^d - d,) - d: {M - U:)~\ ^^M^, (69) 

from which 

I — M^L^s .^^x 

' M4^ {2d - ^,) - d,' (Sd - M,) mT' ^ ^ 

which gives the required width. 



TRANSVERSE STRENGTH. 



55 



Section YI. — Transverse Strength — Circular Sections. 

52. 3Ioment of Resistance. In Fig. 20, let BnOl 
represent the section, BiiOMnNB a section tlirongli the 
strain wedges on the line BO^ B the origin of co-ordinates, 
DBS the axis or origin of moments, nl the neutral line^ nBl 
the compressed area, and n 01 the extended area 




Adopting the notation heretofore used and defined, we 
have 

C = the greatest intensity of the compressive strain in pounds, 

per square inch, 
T = the greatest intensity of the tensile strain in pounds, 

per square inch, 
B^ = the moment of the tensile resistance in inch-pounds, 

7?c = the moment of the compressive resistance in inch- 
pounds, 
B =z the Moment of Kesistance of the section, 
d^ = the versine Bn of one half of the arc nBl in inches, 
d^, = the versine On of one half of the arc nOl in 

inches, 
d =: df,-{- d^, = the diameter of the circle in inches, 
r = the radius of the circle in inches, 
q -— the quotient arising from dividing by T^ 



56 STRENGTH OF BEAMS AND COLUMNS. 

L — tlie total applied load in pounds, 
s = tlie span, the distance between the supports 

in inches, 
m — the factor defined in Art. 34, page 35, 
Comp. arc = the arc bounding the compressed area in 
inches. 
Tension arc = the arc bounding the extended area in inches. 

For the moment of coinjpressive resistance from Eq. 15, 
page 14, Ave have 



K^ = 



C 



21^Z, 



r Vd.d,, [4.d: {d, - r) + r' (30r - Ud,)] 



+ Comp. arc (12<^c — 15r)/ J, (71) 
and for the moment of tensile resistance 



B^ = 



T 

24:d. 



r Vd,d^ [UJ" (5r - d^) + ISr' {r - d,)] 



+ Tension arc {12d^ — 9r) 7-0, (72) 
and for the Moment of Resistance of the circular section, 

B = B, + B, = 2B, = 2E,. (73) 

53, The Neutral Line, The position of the neutral 
line in circular wooden, cast-iron, wrought-iron, and steel beams 
will be found tabulated on pages 80, 100 and 112. 

54, Transverse Strength. Substituting for i?, in 
Eq. 22, page 36, its value 21?, ^ 2R^ from Eqs. 71 and 72, 
we have 

^ = 1^, [ ^^- [^^c^ (^^c - r) + r' {30r - 1^)] 

+ Comj). arc {\2d, - l^r)r'~^, (74) 



TRANSVERSE STRENGTH. 57 

and 



L = 



riiT 



12r4s 



_ [ |/.///, \\d: {^or - cJ;) + 18r^ (r - <,)] 



+ Tension cu^c {V^d^ — 9r) r"" , (75) 

from which the breaking load of a circular beam may be com- 
puted. By observing the following equalities, much time and 
labor will be economized in comparing the results computed 
from these formulas : 



\/d,d^ ]Jtd; (d, - 7') + r' (SOr - Ud,)] 

= VdAl'^dT" (5^^ - d,) + ISr' {r - d^)], 

=F (12fZe - 1-5;") r' = ± (12^, - 9r) r% 

Comp. arc — 27rr — Tension arc. 

Another Method, The Moments of Kesistance of 
circular sections are to each other as the cubes of their rcidii / 
hence by computing and tabulating the Moments of Resist- 
ance for all required jiositions of the neutral line, in a circle 
whose radius is unity ^ those for any other circle composed of 
the same material may be computed by multiplying the 
tabular number by the cube of the radius. 

Let f^ = the second member of Eq. 74, when r = 1^ except 

the factor - — - 

s 

f^. = the second member of Eq. 75, when /» =: 1, except 

m J. 

the factor . 

s 

. • . Z = "^f^, (76) 

and 

L ^ !!?:!5zr. (77) 



58 STRENGTH OF BEAMS AND COLUMNS. 

The application of Eq. 76 is illustrated in Examples 11, 32 
and 33, and Eq. 77 in Examples 12, 13 and 14. 

Tables giving the computed values of /. and /,, for the 
diiferent positions of the neutral line in cast-iron, wrought- 
iron and steel, circular sections, whose radius is unity, are 
given on pages 80 and 100. 

55. To Compute the Compressive Strain. 

Problem. — T/ie position of the neutral line and the com- 
pressive strain of the material of a circular section may he 
computed from the Jcnown transverse hreaking load and tensile 
strength of the material. 

Deducing the value of f from Eq. 77, we have 

from the proper table in the sequel take the value of q cor- 
responding to this value of f ; then 

C=qT. (79) 

Illustrated by Examj)les 3 and 23. 

56. To Compute tlie Tensile Strain, 

Problem. — The tensile strain of the material of a circular 
section may he computed from the known transverse hreak- 
ing load and compressive strength of the material. 

From Eq. 76 deduce the value of /„ 

• • • /c = -— ^7>- (80) 

mr (J ^ 

From the Table given in the sequel for the material take the 
value of q^ corresponding to the computed value of /c^ then 

^ = f (81) 

Illustrated by Example 24. 



TRANSVERSE STRENGTH. 59 

57. Relative Strength of Circular and Square 

Beams, Constant ratios exist between tlie Moments of Ee- 
sistance of the circle and its inscribed, circumscribed and equal 
area square, when the material composing the square and 
the circle has the same tensile and compressive strength, and 
consequently the same ratios exist between the transverse 
strength of the beams of which they are sections, the sj)an and 
manner of loading the circular and square beam being the 
same. 

General Formula : 

Let f = the position of the neutral line in the square when 

the side is unity, and d^ = fd^, h = d and r^ =: -—, 

8 

hence, 

hd ^ O f^d^ C 
Strength of the Square = — ^ — z=''-^- — ■, from Eq. 23. 

Strength of the Circle ■= rfC = ^f^^ f roj^ Eq. 76, 

8 

from which we have 

Square : Circle : : ^^ : ^^, 

^ 3 8' 

. • . Circle = Square X — - - . (82^) 

^ 8^/y'^ ^ ^ 

Case I. — Wlien the circle is inscribed loithin the square^ 
d = d, and Eq. 82 becomes 

Circle = Square X ^^^ (83) 

Illustrated by Examples 15 and 16 of the sequel. 

Case II. — When the square is inscribed within the circle side 
of the square, d = diameter of the circle d X 0.707, and Eq. 
82 becomes 

Circle = Sonar e X • — (84) 

^ 2.8271/2 ^ ^ 



60 



STRENGTH OF BEAMS AND COLUIINS. 



Case III. — WTien the square and circle are equal in area 
side of the square^ d =■ diameter of the circle d X 0.886, and 
Eq. 82 becomes 

*" /^ (85) 



Circle = Square X 



5.564 f 



Illustrated by Examples 18 and 19. 



Section YII. — Transverse Strength — Hollow Circular 

Sections. 

58. Moment of Resistance, In Fig. 21, let BnOl 
represent the section, BnOMnNB a section through the 
strain wedges on the line BO, B the origin of co-ordinates, 
DBS the axis or origin of moments, and nl the neutral line. 




%-2i 



a. 



_ ^./ _ 

— ^^ 



A 



t 

T 



J 



^ -> 



O 



Let t = the thickness of the metal ring in inches, 
r = the radius of the outer circumference. 



r, — r — ^ = 



a a a 



inner 



u 



tC= CsiTid tT= T. 



For other notation refer to Art. 52. 

When t, the thickness of the metal ring, is very small, the 
entire strain distributed over the metal of the section may be 



TRANSVERSE STRENGTH. 61 

treated as if it was all concentrated in the outer surface of the 
cylindrical beam ; if ^ is not very thin, r — ^ or the mean ra- 
dius must be used instead of r in the following formulas. 

Making the substitution of tC for C and tT ior T in Eqs. 
18 and 19, the moments of coTripressive and tensile resistance 
become 

^c = ^[ Vd^, X 2 (3/- - d,) + Comp. arc {2d, - 3r)1, (86) 

and 

J^T = ^1 ^d4^ x2{r^d^)-\- Te7ision arc {2d^— ?') J- ^^'^) 

For the Moment of Resistance of the section, 

E^B,^B, = 2E, = 2^,. (88) 

59, The Neutral Line, The position of the neutral 
line in hollow circular sections of cast-iron, wrought-iron and 
steel, when the radius of the outer circumference is unity, will 
be found tabulated in the sequel. 

60, Transverse Streng^th. Substituting for the 
Moment of Resistance B, in Eq. 22, page 36, its values in this 
case, 2^c = 2^^^ from Eqs. 86 and 87, we have 



Z = 



mrt C 



s dc 
and 



Z = 



mrtT 

S Orr 



[ Vd,d^ X 2 {2>r - d,)+Comp. arc {2d, - 3r) J, (89) 
[^ VdA X2{r + d^)+ Tension arc {2d^— r)\ (90) 



From either of the above formulas the transverse strength of 
hollow cylindrical beams may be computed. 



62 STRENGTH OF BEAMS AND COLUMNS. 

In these formulas the following equalities exist : 



Vd^X 2 (3/' — d^) = Vd^d^ X 2 (/^ + d^\ 

± {2d, -Sr)= T {2d, - r\ 

Oomp. arc = 27rr — Tension arc. 

When the metal ring is very thin, r in the above formulas 
is the radius of the outer circle, otherwise it is the radius of 

the mean circle, r — — . 

Another Method, The Moments of Eesistance of thin 
hollow circular sections are to each other as the squares of 
their radii ; hence, bj computing and tabulating the Moments 
of Resistance for all required positions of the neutral line, in a 
thin hollow circle whose radius is unity, those for any other 
thin hollow circle composed of the same material may be 
computed by multipljdng the tabular number by the square of 
the radius. 

Lety^ = the second member of Eq. 89, when radius = 1, ex- 
cept the factor , 

s 

f^ — the second member of Eq. 90, when radius = 1, ex- 
cept the factor . 

With this notation Eqs. 89 and 90 become 

L = ^^^^, (91) 

s 

and 4 

L = 'Hi'^fil. (92) 



TRANSVERSE STRENGTH. 63 

Tables giving the computed values oif^ and f^. for the posi- 
tions of the neutral line in thin hollow circles, whose radius is 
unity, will be found in the sequel. 

The application of the above formulas is illustrated in 
Examples 20 and 34. 



CHAPTEK lY. 

CAST-IRON BEAMS. 
Section I. — General Conditions. 

61. Compressive Strength, CruMng. — The crush- 
ing strength of cast-iron that is nsually obtained by experi- 
menters and recorded for use in designing structures, is the 
number of pounds avoirdupois that it requires to crush a 
prism of the material whose sectional area is one square inch 
and length from one and a half to three times the diameter, 
under which condition it is found to be more nearly constant 
in value for the same material than when the height bears a 
greater or less ratio to tlie least diameter of the prism tested. 

Yalue. — The range of values for the crushing strength of 
cast-iron may be taken as being from 85000 pounds to 125000 
pounds per square inch ; the mean is al)0ut 100000 pounds. 

Elastic Limit. — The compressive elasticity of cast-iron as 
recorded by the earlier experimenters appears to be very 
defective, but improved methods of manufacture have pro- 
duced a cast-iron from which modern experimenters find the 
increase in the amount of the compression to be, practically, 
in direct proportion to the increase in the load, within the 
elastic limit of the cast-iron. 

The compressive elastic limit varies from three fifths to 
nearly the crushing strength. 

63. Tensile Streiijj^th. T'<??^,ac^Yy/ is the force in pounds 
that is required to pull asunder a prism of cast-iron whose sec- 
tional area is one square inch. It ranges in value from 15000 
to 3000.0 pounds. 



GENEIiAL CONDITIONS. 65 

Elastic Limit— T\\Q extensions Avitliin the elastic limit fol- 
low laws similar to those that govern the compressions. The 
elastic limit is about one half of the tensile strength. 

63, Ratio of the Compressive to the Tensile 
Streiig-th. Experiments have demonstrated that the ratio 
existing between the crushing and tensile strength of cast-iron 
has a wider range of values than that for any other known 
material, and that this results from great fluctuations in the 
crushing rather than in the tensile strength. The extreme 
values for the C ^ T = q iyi our formulas may be taken at 
from 3 to 8f ; the tendency of improved methods of manu- 
facture is to decrease, numerically, this ratio by increasing 
that of rand decreasing that of L\ which in pure or wrought- 
iron becomes C ~ T = \. 

64. Transverse Streng^th. Cast-iron breaking with 
a well-deflned fracture, its transverse strength may be com- 
puted from our formulas when the crushing '^.l^^ tensile 
strength of ih^ identical cast-iron composing the beam is 
known from experiment, as the values of C and T are found 
to vary in an uncertain manner, with remelting, length of 
time in fusion, etc. 

^ The tranmerse strength of some cast-iron does not increase 
directly with the increase in the dimensions of its cross-section, 
as it should in accordance with well-defined laws, but in a 
lower ratio. This defect is usually imputed to unequal strains 
being brought upon the metal in different parts of the section, 
from unequal temperatures in cooling, but few experiments 
have been made to test the matter. 

Col. James, of England, planed out and tested a |-inch 
square bar from a 2-inch square cast bar, the values of C and 
r for this brand of iron, Clyde ISTo. 3, having' been determined 
from experiments by Mr. Eaton Ilodgkinson. The following 
is the result of Col. James's tests : 



66 STEEN^GTH OF BEAMS AND COLUMNS 

Clyde No. 3. The Dressed Bar. Transverse Load. 

(7 = 106039 lbs. C = 60233 lbs. by test. 193 lbs. rough bar. 

T= 23468 "• r = 14509 " by computation.* 134 '' planed out. 

From the above it appears that tlie vahie of C decreased in 
tlie "planed-out" bar 56.8 per cent; that of T 60.5 per cent, 
and the transverse breaking load 69.4 per cent. But this 
result is contrary to that obtained in the following tests made 
for the United States Government — Rej)ort of Tests of Iron and 
Steel for 1885.t In the United States Government tests the 
bars were cast two feet in length and three inches in width, 
with depths ranging from a half to two inches. The equiva- 
lent centre breaking load for a bar of the metal one inch 
square and twelve inches span was computed from the result 
of each experiment, that they might be compared. 

Condition. Rough. Edges Planed. Edges Planed. Planed all over. 

No. Expts. 4 3 6 6 

Size, 3" X 0".5 2" X 1" to 1".28 2" X 1".5 to 2" 2" X 1".25 to 1".75 

Load, lbs., 2453 2136 2025 2134 

The " edges planed" bars were reduced in width from the 
rough castings that were 3" wide, the tension and compression 
vsurfaces being left as they were cast. 

The " planed-all-over" bars were cut from cast bars that 
were three inches wide and two inches deep, equal depths of 
metal having been removed from each face of the rough bar. 

In these experiments the strength of the bars that were cast 
two inches deep varied about live per cent from tliose cast 
one inch deep, and the " planed-out" bars were as strong as 
the rough cast bars of the same size. 

The bars that were tested with the tension and compres- 
sion surfaces as cast varied in strength about 5 per cent from 
the average strength given al)ove, and the " planed-out " bars 



* Example No. 4. 

f Senate Ex. Doc. No. 36— 49th Congress, Ist Session, p. 1162. 



GEJS^ERAL CONDITIONS. 67 

only 2J per cent in strength from the averages given, indicat- 
ing great uniformity in strength, and consequently no de- 
crease in the vahies of C and T for the metal in the interior 
of the cast bar, but this was not determined by direct exper- 
iment. 

The following ratios are usually quoted to illustrate the de- 
preciation of the transverse strength of cast-iron beams re- 
sulting from an increase in the size of the cross-section, the 
strength that the large beam should have being computed 
from that of the \" X V bars, the latter beincr denoted by 
100. '^ 

Experimenter. 1" X 1" 2" X 2" 3 " x 3" 

Capt. James loo 71.84 61.95 per cent, 

Mr. Hodgkinson . . 100 .71.22 <-- 

The transverse elastic limit load may be computed when 
the corresponding values of C and T have been determined 
by experiment. 

65. To Compute the Compressive and Tensile 
Strength. The relation existing between the Transverse 
Load, Compressive and Tensile Strength is such that the 
value of any one of the three may be" computed when the 
value of tlie other two has been determined by direct experi- 
ment ; this relation is true for both the elastic limit and the 
hreaJcing values. 

The beam from which the experimental breaking load was 
obtained must have been sufficiently long to deflect a distance 
equal to or greater than the depth of tension area that is re- 
quired iovtrue cross-breaking, or else our computed value of 
C or T will be incorrect for the reason given in Art. 24. 

Com,pressive Strength.— Thh may be computed from the 
known tensile strength of the material and the transverse 
breakmg load of rectangular and circular beams, or of that 
of any of the flanged beams when the neutral line lies within 
the lower or tension flange. 



68 STRENGTH OF BEAMS AND COLUMNS. 

Tlie cast-iron that Mr. Barlow used in his experiments was 
composed of pig and scrap-iron ; lie determined its tensile and 
transverse strength, but he does not record in his " Strength 
of Materials " its crushing strength ; this we can compute from 
the data given. 

ExA^iPLE 1.— Kequired the Crushing Strength of Mr. Bar- 
low's iron from the centre-breaking transverse load of a rect- 
angular beam when 

The depth ..d^l inch, T = 18Y50 pounds by experiment, 
'' breadth h = 1.02 inch, Z = 531 " " " ^ 

" span . . . 5 = 60. inches, m = 4. 

The position of the neutral line must be computed from 
Eq. 30, page 40, 

_ 3X1 _ /9 X 4 X l.UJi(l)^ 18750 - 12 X 534 X «0 ^ q„ ^^^^ 
^^ - 2 V 4 X 4 X 1.02 X 18750 

hence ^c = 1 - 0.5032 = 0".4968, 

from w^hich we can compute the value of the crushing strain 

by means of Eq. 31, 

C= _3_>^3J:_X_60 ^ pounds. 

4 X 1.02 (0.4968)^ ^ 

The following Example is taken from Major Wade's Ex- 
periments on the " Strength and other Properties of Metals 
for Cannon," made for the United States Government : 

Example 2.— Kequired the Crushing Strength of Major 
Wade's cast-iron in third fusion, from the transverse strength 
of a rectangular beam, when 

The depth. . .d = 2.01 inches, T = 26569 pounds by test, 
'^ breadth h = 2.008 " Z = 16172 '' " " 
" span 6' == 20. " m = 4 for a centre load. 



* Barlow's " Strensrth of Materials," p. 152. 



GENERAL CONDITIONS. 69 

For the position of the neutral line, we have from Eq. 30, 

_ IX?:^ _ /9 X 4x 2.008(2.0iF26569^ 12 X 16172 X 20 _ ^„ ^^^^ 
^-2 y 4 X 4 X 2.008 X 2¥569 " ^ '^^^^' 

. • . ^e = 2.01 - 0.8835 = 1.1265 inches. 

Then we can compute the crushing strength, O, from Eq. 
31, 

„ 3 X 16172 X 20 ^._^ , 



Example 3. — Required the Crushing Strength of certain 
,st-ir 
when 



cast-iron from the transverse strength of a circular beam 



The diameter d = l'M29, T= 29400 pounds mean of tests, 
'• span s = 20'^0, Z = 2118 " by test, 

771 = 4. 

From Eq. Y8 we have 

^ 2118 X 20 ^ 

-^^ 4 (0.5645/ 29400 

From the Table, page 80, for the above value of /t> we have 
q = 3.275, and from Eq. 79, 

. • . 0= 29400 X 3.275 = 96285 pounds. 

From the United States Government Report of the Tests of 
Iron and Steel for 1884,"^ the mean crushing strength of 
this cast-iron was C — 100700 pounds. 

The C^mshing Strength may be computed from the Trans- 
verse Strength of the Hodgkinson or any other flanged beam. 
When the neutral line is not situated above the top line of the 
tension flange, its position may be computed from Eq. 30, 
page 40, as if the beam were rectangular. When the neutral 



* Senate Ex. Doc. No. 35— 49th Congress, 1st Session, p. 284. 



70 STRENGTH OF BEAMS AND COLUMNS. 

line is located above the top line of the tension flange, its 
position can be determined by the method given in Problem I., 
page 43, but in either case the value of C must be computed 
bj formula 37, page 4-1:, C being then the only unknown 
quantity that it will contain. 

Tensile Strength. — This may be computed from the known 
Crushing Strength of the material and the Transverse 
Strength of any rectangular, circular or flanged beam. 

Example 4. — Required the Tensile Strength from the Trans- 
verse Strength of a rectangular beam made of a certain kind 
of cast-iron that was tested by Captain James, of England,* 
when 

The depth d = 0.75 inches, C= 60233 pounds by test, 
" breadth h = 0.75 " Z = 134 " " • " 

" span s = 54.0 " fn = 4: 

The position of the neutral line is to be computed from Eq. 

32, page 40, 

. ,/ 3X134X54 r. ^ ' I 

^^ = ^' 4-^0:75-^^60233 = '-^ ^^^^^^^^' 

hence d^, = 0.75 — 0.4 = 0.35 inches. 

Then we compute the required value of the tensile strength, 
Ty from Eq. 33, page 40, 

^ = 1 TTl^ TTTi^VT^ r. ^K ■ ^ -T^ = 14509 DOUuds. 

4 X 0.75 X 0.35 (2 X 0.75 + 0.4) ^ 

Mr. Hodgkinson determined for this iron, 

Clyde, ISTo. 3, O = 106039 and T = 23468 pounds. 



* British " Report on the Application of Iron to Railway Structures," 
p. 257. 



EECTANGULAR CAST-IRON BEAMS. 



71 



Section II. — Rectangular Cast-iron Beams. 

66. The Neutral Line. The position of the neutral 
hne in rectangular cast-iron beams, for the different required 
ratios of C ^ T — q^ have been computed from Eq. 2(3, page 
38, and tabulated below for reference. 

Table of positions of the neutral line in rectangidar cast- 
iron heams. 



Ratio of Crushing 


1 

I Depth of Neutral Line 


Ratio of Crushing 


Depth of Neutral Line 


to Tenacity 


Below the Crushed Side 


to Tenacity, 


Below the Crushed Side 


or 


of the Beam, or 


or 


of the Beam, or 


C-5- T^ q. 


dc. 


C-f- T=q. 


dc. 


8.0 


0.4191 d 


5.5 


0.4831 d 


7.875 


0.4127 d 


5 . 375 


0.4871 d 


7.75 


0.4243 d 


5.25 


0.4913 d 


7.625 


0.4226 d 


5.125 


0.4956 d 


7.5 


0.4298 d 


5.0 


0.5000 d 


7 375 


4326 d 


4.875 


0.5045 d 


7.25 


0.4354 d 


4.75 


0.5092 d 


7.125 


0.4384 d 


4.625 


5139 d 


7.0 


0.4414 d 


4.5 


5189 d 


6.875 


0.4444 d 


4.375 


5240 d 


6.75 


0.4475 d 


4.25 


0.5293 d 


6.625 


0.4507 d 


4.125 


0.5347 d 


6.5 


0.4540 d 


4.0 


0.5403 d 


6.375 


0.4573 d 


3.875 


0.5461 d 


6.25 


0.4608 d 


3.75 


0.5521 d 


6.125 


0.4642 d 


3.625 


0.5583 d 


6.0 


0.4678 d 


3.5 


5647 d 


5.875 


0.4715 d 


3.375 


0.5714 d 


5.75 


0.4754 (/ 


3.25 


0.5784 d 


5.625 


0.4791 d 


3.125 


0.5855 d 






3.0 


0.5930 d 



Should lower numerical values of q be required, refer to the 
table of neutral lines in rectangular Sections of Steel. 



67. Transverse Streng-th, From either Eq. 27 or 
28, as may be most convenient, the transverse strength of rect- 
angular cast-iron beams may be computed. 



72 STRENGTH OF BEAMS AND COLUMNS. 

Example 5. — Required tlie centre breaking load of a rect- 
angular cast-iron beam when 

Tlie dej^tli d — 1 incli, C = 10243i pounds by test, 

" breadth l =. 1 '^ 7^=10724: " " " 

" span 6' == 54: inches, ^ = 6 . 125 and ??^ = 4. 

The position of the neutral line may be computed from 
Eq. 26, 

. 1(- 0.5-fi/2x 6.125+2.25) ,, ,^,,, . -, 
< ^ r, ,0- I 1 -^- = 0.4642 mches, 

or its value may be taken from the Table. 

The transverse breaking load, Z, from Eq. 28, becomes 

^ ^ 4 X 1(0.4642)- X102J34 ^ ^^, ^^^^^^ 
3 X 54 ^ 




Mr. Hodgkinson, in his experiments, 
gives the above values of C and T for 
Blaen-Avon No. 2 iron, and 447 pounds 
as the mean transverse streno^th of 4 
beams tested, wliile Captain James found 
556 pounds to be the mean of 3 beams 
of the above dimensions for this brand of 
iron. 



Example 6.— Required the centre breaking load of the fol- 
lowing : 1 inch square cast-iron beams, the span being 54 
inches, and the values of C^ 7" and the experimental break- 
ing loads being from Mr. Hodgkinson' s experiments. It is 
very probable tlmt tliese bars did not deflect sufficiently to de- 
velop the true transverse strength.'" 



* Barlow's " Strengtli of Muteriuls," p. 163. 



RECTANGULAR CAST-IROJS^ BEAMS. 



73 







Tested Strength per 




Transverse 






Square 


Inch. 


Neutral 

Line, or 

dc. 


Strength. 






Crushing C. 


Tensile T. 


Computed, 
lbs. 


Tested. 












lbs. 


CaiTon No 


2, C B.... 


106375 


16683 


0.457 


548 


476 


( ( ( < 


2, H. B. . . 


108540 


13505 


0.418 


468 


463 


i( te 


3, C. B.... 


115442 


14200 


0.416 


493 


446 


" " 


3, H. B. . . 


133440 


17755 


0.427 


606 


527 


Buffery " 


1, C. B.... 


93366 


17466 


0.488 


549 


463 


ii ( ( 


1, H. B. . . 


86397 


13434 


0.456 


429 


436 


Mean . . . 






515 


468 







Example T. — Required tlie centre breaking load of a rect- 
angular cast-iron beam, from Mr. Barlow's experiments, 
when 

The depth ..d= 2.00 inches, C = 95462 lbs., computed, Ex. 1, 
" breadth h = 1.99 " T= 18750 " by test, 
" span. . . ..9 = 60.00 " ^ = 5 . 091 and m = 4. 

For the neutral line we have, from Eq. 26, d^ = 0''. 993-1, 

Computed transverse strength from Eq. 27 or 28 . . . .4175 lbs., 
Tested " " 3863 " 

68. To Design a Rectangular Cast-iron Beam. 

The principles and formulas required are given in Art. 38, 
page 39. 



74 STRENGTH OF BEAMS AND COLUMNS. 



Section III. — Ilodghinson Cast-iron Beains. 

69. In this section we will consider the principles as if 
applied to that class of cast-iron beams whose bottom or 
tension flange is larger than its top or compressed flange, 
though they are applicable to all other forms of flanged beams. 

70. Mr. Hodgkinson's Experiments, Previous 
to the year 181:0, when Mr. Hodgkinson commenced his ex- 
perimental investigation into the strength of materials, the 
only metal beam that had been used in structures to any ex- 
tent was the Inverted T, Fig. IT, and the Double T, Fig. 1(), 
in which the top and bottom flanges Avere equal in area. 
Commencing his experiments with the equal flanged Double 
T, he found that, by increasing the area of the lower or tension 
flange by small amounts, he continued to increase the 
transverse strength of the beam, his unit of measure and 
standard of comparison being the quotient obtained from di- 
viding the breaking load by the area of the section of the 
beam expressed in inches, and that this continued to increase 
until he had reached the point where the tension flange was 
about six times the area of the compressed flange ; increasing- 
it to a greater ratio he found that the transverse strength j^er 
square inch of section began to decrease. Experimenting with 
cast-iron, in which the ratio of C to jT, or the crushing to the 
tensile strength, was about six^ he recommended that the ten- 
sion flange should have six times the area of the compressed 
flange, in order to obtain the greatest transverse strength per 
square inch of section. But subsequent investigators, experi- 
menting with cast-iron possessing greater tenacity^ or a less 
ratio of (7 to 7^ than that used bj^ Mr. Hodgkinson, found the 
greatest strength with a lower ratio of extended to compressed 
flange than that recommended by him. The reason for this 
will be apparent when the formulas for the strength of these 
beams are examined. 



1 ' 1 

.\/ . . 


_ b. _■ 

b 


; 2 


Z 1 



HODGKINSON CAST-IRON BEAMS. 75 

71. Neutral Line. When the neutral Kne is not situ- 
ated witliin the tension flanges, its position may be ajjjjwxi- 
mately computed from Eq. 3G, page 42. Should the neutral 
line lie witliin the tension flange, its position may IJe accu- 
rately computed from Eq. 30, or from the methods given in 
Probs. 1 and 2, page 43, when the transverse load and either 
the compressive or tensile strength has been determined from 

experiments. 

AFT 1 

73. Transverse Strength. The 

transverse strength of the Hodgkinson or 
of any beam having flanges at Xh^ top 
and the bottom may be computed from 
either Eq. 37 or 38, as may be most con- 
venient. 

ExA^MPLE 8. — Required the centre breaking load of a Hodg- 
kinson cast-iron beam when 

The depth of the beam ^ = 5.125 ins., C = 96000 lbs., 

" breadth of the Aveb J = 0.34 '' T = 16000 " 

" " " top flanges, h, = 1.22 " q= 6 

" depth '' " "' d, = 0.315 " s = U ins. 

" '' bottom " d,= 0.56 '' m = 4. 

'' breadth " " '' h, — 4.83 " 

Nexdral Line. — Our formula 36, giving the position of the 
neutral line approximately only, places it in each case a little 
too near the tension flange. In this example it locates it 
within the tension flange, and we will assume that it coin- 
cides with its upper line, or 

d^ = 4^565 and d^ = 0.56. 

Erom Eq. 37 the transverse breaking load becomes 

J- _ 0.34 (4.565)3+1.22 (0.315)^ (3X4.565 -2X0.B15^ .vQfinon i-«ifiiy.. 
J^ - 3X4565X54 ' ^X^^OOO = 1,616 lbs. 

Tlie tension area beiner continuous from the neutral line to the 



76 STRENGTH OF BEAMS AND COLUMNS. 

bottom of the beam, Eq. 28, page 38, can be used to compute 
the load from the tensile strength T, in wliich h ^ h -\- h^ oi 
this example. 

-r 4x5.17x0.56^2x5.125+4.565)16000 _,^^,,, „ 

L = • ^ — z~, ' — 16944 lbs. 

3Xo4 

These two values of Z, not being identical in numerical value, 
our assumed position for the neutral line must liave been a 
little too low down in the section for exact equilibrium. 

The above described beam was one of the series of beams 
that was used by Mr. Hodgkinson to determine the " section 
of greatest strength." Its transverse breaking load was, by 
test, 16730 pounds." The value for T that we use was also 
determined from experiment ; the ratio was supposed to be 
6^ -^ T = 6. In this series of beams the neutral line of 
rupture "was continuously lowered in the section, from pro- 
gressively increasing the area of the tension flange until it 
finally reached the upper surface of this flange. 

Example 9. — Required the centre breaking load of a Hodg- 
kinson Cast-iron Beam (Fig. 15), when 

The depth of the beam d = 14.0 ins., C = 75983 lbs. by test, 

" breadth of the web b= 1.0" 7=13815" " " 

top flano:es di= 1.0" ^ = 5.5, 

"depth " " " b, = 2.5" ,s = 16 feet, 

" " " bottom flanges . ..^2 = 1.75" m = 4. 

" breadth " " " ...M= 11.00" 

From Eq. 36 the position of the neutral line becomes 

, _ 1X14+ Vf2xl4xllXl.7 5(5.5 + l)4-(lyl4)M4vo 5+5) _,.,_. 
"^^ - 2x1(5.5+1) - ^'^'^ '"'" 

which is within the tension flanges, assuming d^ := 14 — 1.75 == 
12.25. 

* Barlow's " Strength of Materials," p. 177. 



HODGKINSON CAST-IROI^^ BEAMS. 77 

From Eq. 37 the breaking load, Z, becomes 

_ 1(12.25)3 + 2.5 (1 )^3x12.25-2x1) _ 

^-^ - 3 X 12725 X 12 X 16 ^ ^ ^ '^^ ~ ^'^^ " • 

From the tensile strength, T^ and Eq. 28, L becomes 

^ 4 X 12 X 1.75 (2 X 14 + 12.25) 13815 ^^^. ^„ ,, 

X = — ^ — , — sUlOy lbs. 

3 X 12 X 16 

From this near identity of values for L we conclude that our 
assumed position for the neutral line was very near the correct 
position. 

Two of these beams w^ere constructed of Calder No. 1 cast- 
iron, and broken with centre loads of 73920 and 76160 pounds 
respectively by Mr. Owens, Inspector of Metals for the British 
Government."^' The values of G and 7^ were determined by 
Mr. Hodgkinson for this brand of iron. The bottom table 
contained six times the area of the top tahle^ as recommended 
by Mr. Hodgkinson. 

These beams broke with the deflections 1^^87 and 2''.0 re- 
spectively, and the full strength of the section was thus 
developed by the transverse load. 

The following statement gives the proportion of the load 
sustained by each uneinher of the transverse section and the 
proportion of the area that it occupied : 

Load. Area, 

Compressed Flanges 2.3 7.0 per cent. 

" Web 47.7 34.3 " 

Extended '" : 4.2 5.0 '' 

'' Flanges 45.8 53.7 " 

100.0 100.0 

The web extends through the total depth of the beam. 

* Box's " Strength of Materials," p. 203. 

4 



78 STRENGTH OF BEAMS AND COLUMNS. 

73. To Design a Hodgkiiison Beam. The 

formula required in designing a IJodgkinson beam will be 
found in Art. 42, page 44. 



Section IV. — Douhle T and Box Cast-iron Beams. 

74. The Neutral Line. The position of the neutral 
line may be approximately computed from Eq. 36, page 42, 
by giving the letters of tlie formula their definitions in Art. 
44. 

75, Transverse Strength. The Transverse Strength 

^ , ^-vj. may be' computed from Eqs. 37 and 38, 

I ' 1 • ^ I giving the letters their definition in Art. 



(-- 



j( 



i,_ 



"& 



44 
I 

X Example 10. — Required the centre 

— "|i— , ^ breaking load of a Hollow Rectangular 



"^ I Cast-iron Beam whose outside dimension 



is 3';i25 X 3".125, inside 2".375 X 
2".375, thickness of metal all around 0.375, when 

The depth of the beam. .....d = 3'M25 C = 84000 lbs. 

" breadth '' sides b = 2 X 0".375 T= 14000 '' 

" top b, = 2^375 q =Q, 

'' depth '' " ^. =:. 0'^375 ^ == 6 f eet, 

" " " bottom .. .r/, = 0".375 m == 4. 

" breadth '' " . ..^, = 2^^375 

The position of the neutral line d^ from Eq. 36 becomes . 

dc = 

3.125X0.754- 4/12x3.125 X 0-75 X 2.37 5 (6+1)4- (3.125+0.75)' (4x6+5 )= 

' ' 2X0.75 (6 + 1) 

1".965. 



DOUBLE T AND BOX CAST-IRON BEAMS. 79 

The breaking load from Eq. 37 becomes 

, 0. 75 (1.965)» +2.375 (0.375)^3X1.965-2X0.375) , 
L= ^3^ 965X12X6 X4x84000=5760 lbs., 

and from the tensile strength 7", and Eq. 38, L becomes 

L = 5178 pounds. 

Tn the investigations preliminary to the construction of the 
Menai Straits Tubular Bridge, Mr. Stephenson broke this 
beam with 5387 pounds.^ 

76. To Design a Double T and Hollow Rectangular Cast- 
iron Beam. This may be done by using the formula and 
directions given in Art. 47, page 48. 

* " Britannia and Conway Tubular Bridges," p. 429. 



80 



STRENGTH OF BEAMS AND COLUMNS. 



Section Y. — Circular Cast-iron Beams. 

7 7 . Neutral Line. Tahle of positions of the neutral line 
in Circular Cast-Iron Beams and factors for use in Eqs. 76 and 77. 



Ratio of Crushing 
to Tenacity, or 



Depth of Nentral 

Line Below the 

Crushed Side of the 

Beam, or dc. 



8. 

7.875 
7.75 
•7.625 
7.5 
7.375 
7.25 
7.125 
7.0. 
6.875 
6.75 
6.625 
6.5 
6.375 
6.25 • 
6.125 
6.0 
5.875 
5.75 
5.625 
5.5 
5.375 
5.25 
5.125 
5.0 
4.875 
4.75 
4.625 
4.5 
4.375 
4.25 
4.125 
4.0 
3.886 
3.75 
3.625 
3.5 
3.375 
3.25 
3.125 
3.0 



0.3984 d 

0.4004 d 

0.4022 

0.4041 

4062 

0.4087 

0.4112 

0.4137 

0.4161 „ 

0.418^ d 

0.4201 d 

0.4236 

0.4262 

0.4289 

0.4316 

0.4344 

0.4373 

0.4402 

0.4432 

0.4463 

0.4494 

0.4526 

0.4559 

0.4594 

0.4629 

0.4665 

0.4702 

0.4740 

0.4780 d 

0.4821 d 

0.4863 d 

0.4906 d 

0.4951 

0.5000 

0.5045 

0.5095 

0.5146 

0.5200 d 

0.5256 d 

5314 d 

0.5375 d 



Factors for Computing the Moment 
OF Resistance, r = 1. 



fc for Crushing 
Strain, C. 



0.3226 
0.3260 
0.3298 
0.3336 
0.3380 
0.3426 
0.3472 
0.3518 
0.3568 
0.3620 
0.3672 
0.3726 
0.3780 
3836 
0.3892 
0.3952 
0.4014 
0.4078 
0.4142 
0.4210 
0.4278 
0.4350 
0.4426 
0.4504 
0.4584 
0.4668 
0.4752 
0.4842 
0.4938 
0.5036 
0.5138 
0.5244 
0.5356 
0.5478 
5593 
0.5722 
0.5855 
0.5996 
6145 
0.6302 
6468 



/V for Tensile 
Strain, T. 



2.5804 

2.5722 

2.5640 

2.5556 

2.5466 

2.5356 

2 5250 

2 5138 

2.5030 

2.4920 

2.4808 

2.4694 

2.4578 

2.4458 

2.4346 

2.4208 

2.4080 

2.3952 

2.3812 

2.3664 

2.3536 

2.3392 

2.3240 

2.3080 

2.2916 

2 2750 

2 2580 

2.2400 

2.2218 

2.2028 

2.1832 

2.1630 

2.1424 

2.1186 

2.0974 

2.0739 

2.0493 

2.0237 

1.9970 

1.9691 

1.9399 




CIRCULAR CAST-IRO]^^■ BEAMS. 81 

From this Table the position of the neutral line is obtained 
by placing for cl^ the diameter in inches 
and computing the product indicated. 

•78. The Transverse Streiig:tli 

of Circular Cast-iron Beams may be ac- ^* 
curately computed from either Eq. 74 or 

75, but as this involves a tedious calcu- 

lation Eq. 76 or 77 will give results, practi.'.allv exact, with 
much less labor. 

Example 11.— Eequired the centre breaking load of a Cir- 
cular Cast-iron Beam, when 

The diameter .7 = 2^0, 6^ = 95200 pounds computed, Ex 2 

" ^Pa« '^ = 20^0, T = 26569 " by test, 

m =4, q = 3.58. 

The values of the factors /, and/,, obtained by interpolation 
Irom the Table, page 80, and substituted in Eqs. 76 and 77, 
give 

^_4(l)n).577x 95200 ,_ ' 

^ "20 "" ^^^^^' pounds. 

Major Wade, in his experiments, broke this beam with 
11112 pounds. 

Example 12.— Eequired the centre breaking load of a Cir- 
cular Cast-Iron Beam when 

The diameter.. .,/ = 2^42, C = 95200 lbs. computed, Ex. 2, 

^P^^ -^ = 20^0, T = 26569 - bv test, 

m = 4, q = 3.58. 

/. = 2.065 from the Table, which, substituted in Ecp 77, gives 

^_4(1.21)='2.065x 26569 , 

^ o?) = 1^^^^ pounds. 



82 STRENGTH OF BEAMS AND COLUMNS. 

Major Wade broke four beams of the above dimensions with 
18141, 20419, 19997 and 18225 pounds respectively, the 
mean strength being 19198 pounds. 

Example 13. — Required the centre transverse elastic liinit 
load of a Circular Cast-Iron Beam, when 

The diameter d ■= 1M29, C = 20000 pounds by test, 

" span s = 20".0, T = 17000 

m = 4, q = 1.111. 

The value of ^/t = 1.1818 from the Table, page- 80, substi- 
tuted in Eq. 77, gives 

^ 4 (0.5645y 1.1818 X 17000 . ^^^ , 

X = — ■= IOC) 2 pounds. 

The elastic limit load, of the two of these beams that were 
tested, was 1130 pounds each, the values of 6'' and T being the 
mean of three tests, as given in the United States Government 
Eeport of the Tests of Iron and Steel for 1884."^ 

Example 14. — Kequired the centre breaking load of the 
Circular Cast-Iron Beam whose elastic Ihnit load was com- 
puted in Example 13, when 

> 

C = 100700 pounds by test, /,. = 2.0333 from the Table, 

T= 29400 " " ^=3.422. 

Erom Eq. 77 we have 

J- 4 (0.5645)' 2.0333 X 29400 

X = — ^ ~ = 2131 pounds. 

20 ^ 

The breaking load of the two beams described in Exampk^ 
13 was 2118 and 1795 pounds respectively. 

79. Movement of the Neutral Line. In the Cir- 
cular Cast-iron Beam, Example 13, the depth of the neutral 

* Senate Ex. Doc. No. 35 -49th Congress, 1st Session, p. 284. 



CIRCULAR CAST-IRON BEAMS. 83 

line below tlie compressed surface at the elastic limit was, 
from the Table, page 80, d^ = 0".7853 in Example 14, at the 
instant of rupture, d^ — 0".5846 ; hence, as the loading pro- 
gressed, the neutral line moved upvmrd, or from the tension 
side toward the compressed side of the beam. 

80. Relative Strength of Circular and Square 
Cast-iron Beams. 

Case I. — When the circle is inscribed within the square. 
The relation required is given in Eq. 83. 

Example 15. — Required the relation between the transverse 
strength of a square and the inscribed circular cast-iron beam 
when 6^ -=- 7^ = 5. 

f = 0.5 from Table, page 71, /"c = 0.4581: from Table, page 80. 

3 X /c 3 X 0.4584 ^ „„_ 

. • . Strength of Circle = Strength of the Square X 0.6876. 

Example 16. — Required the centre breaking transverse load 
of the circular beam inscribed within the square, from the 
tested strength of the square beam, wdien 

The side of the square d = 2". 01, q = 3.58, 
" diameter of the circle fZ= 2".01,/e = 0.577 from the Table, 
"span ^ =: 20^0,/ .=: 0.561 " " 

With these values Eq. 83 becomes 

3 X/e ^ 3 X 0.577 ^ ^ 
8 X /' 8 (0.561)"^ 

Actual b'k'g load square beam, Maj. AVade's tests, 16172 lbs. 

" " '' circular " " " " 11112 " 

Comp'd " " '' " =16172X0.687=11110 " 

Case H. — When the square is inscribed tvlthin the circle. 

The relation will be given by Eq. 84. 

Example 17. — Required the relation between the strength 



84 STRENGTH OF BEAMS AND COLUMNS. 

of the circular and its inscribed square beam, when C -^ T r= ^. 

f = 0.5 from the Table, page 71, /c = 0.4584 from the Table, 

page 80. 

. 3 /; ^ 3 X 0.4584 ^ 

* * 2.8271 /•■' 2.8271 (0.5/ * ' 

. • . Sti'ength of the Circle — Strength of the Square X 1.945. 

Case III. — When the circular is equal in area to that of 
the square heam. 

The relation will be given bj Eq. 85. 

Example 18. — Required the relation between the transverse 
strength of the circular and the square beam, when their areas 
are equal and 6^ -^ 7^ = 5. 

/ = 0.5 from the Table, page 71,/e = 0.4584 from the Table, 
page 80. 

, 3/e ^ 3X0.4584 _^ 
' ' 5.564/^ " 5.564 X (0.5/ " ' 

. • . Strength of the Circle = Strength of the Square X 0.987. 

Example 19. — Required the centre breaking load of a Circu- 
lar Cast-Iron Beam from that of the square beam of equal area, 
when 

The side of the square d- 1".01, g = 5.091, 

" diameter of the circle <^ = l".145,/e= 0.4562, from the Table, page 80, 
" span.. s=60".0 /= 0.496 " " " " 71. 



With these values Eq. 85 becomes 

3 X 0.4562 



= 1. 



5.564 (0.496/ 

Mean b'k'g I'd of 4 of these square beams from Mr. Barlow's tests,* 519 lbs., 

" " of the circular beam from Mr. Barlow's tests, 519 pounds, 
Comp'd " " " " " = 519 X 1 = 519 pounds. 



* Barlow's "Strength of Materials," p. 153. 



HOLLOW CIRCULAR CAST-IRON BEAMS. 



85 



Section YI. — Hollow Circular Cast-iron Beams. 
81. Neutral Line. 







Factors for Computing the Moment 


Ratio of Crushing 
to Tenacity, or 


Depth of Neutral 

Line Below the 

Crushed Side of the 

Beam, or rfc- 


op Resistance, r = 1. 


V-^ T^q. 


/c for Crushing 


/t for Tensile 






Strain, C. 


Strain, T. 


8.319 


0.5000 d 


0.8584 


7.1414 


8.25 


0.5020 d 


0.8640 


7.1282 


8.125 


0.5057 d 


0.8742 


7.1104 


8.0 


0.5093 d 


0.8852 


7.1820 


7.875 


0.5131 d 


0.8962 


7.0580 


7.75 


0.5170 d 


0.9076 


7.0352 


7.625 


0.52U9 d 


0.9190 


7.0078 


7.5 


0.5249 d 


0.9304 


6.9818 


7.375 


0.5290 d 


0.9430 


6.9550 


7.25 


0.5331 d 


0.9556 


6.9278 


7.125 


0.5374 d 


0.9684 


6.8996 


7.0 


0.5418 d 


0.9816 


6.8704 


6.875 


0.5462 d 


0.9950 


6.8410 


6.75 


0.5507 d 


1.0090 


6.8104 


6.625 


0.5553 d 


1.0232 


6.7790 


6.5 


0.5600 d 


1.0380 


6.7470 


6.375 


0.5647 d 


1.0534 


6.7174 


6.25 


0.5697 d 


1.0688 


6.6800 


6.125 


0.5748 d 


1.0848 


6.6442 


6.0 


0.5800 d 


1.1008 


6.6092 


5.875 


0.5852 d 


1.1184 


6.5694 


5 75 


0..5907 d 


1.1356 


6.5296 


5.625 


0.5962 d 


1.1540 


6.4918 


5.5 


0.6018 d 


1.1730 


6.4492 


5.375 


0.6076 d 


1.1924 


6.4062 


5.25 


0.6134 d 


1.2124 


6.3640 


5.125 


0.6193 d 


1.2330 


6.3184 


5.0 


0.6255 c? 


1.2542 


6.2702 


4.875 


0.6314 d 


1.2748 


6.2266 


4.75 


0.6384 <^ 


1.2994 


6.1718 


4.625 


0.6451 d 


1.3230 


6.1192 


4.5 


0.6520 ^ 


1.3482 


6.0638 


4.375 


0.6590 <^ 


1.3740 


6.0064 


4.25 


0.6662 d 


1.4000 


5.9492 


4.125 


0.6735 <? 


1.4274 


5.8877 


4.0 


0.6811 d 


1.4559 


5.8240 


3.875 


0.6889 (Z 


1.4856 


5.7569 


3.75 


0.6970 (Z • 


1.5168 


5.6890 


3.625 


0.7052 d 


1.5488 


5.6148 


3.5 


0.7137 cZ 


1.5826 


5.5387 


3.375 


0.7223 d 


1.6172 


5.4578 


3.25 


0.7313 (^ 


1.6538 


5.3745 


3.125 


0.7405 d 


1.6922 


5.2880 


3.0 


0.7500 d 


1.7320 


5.1961 



86 



STRENGTH OF BEAMS AND COLUMNS. 



The above Table gives the positions of the neutral line in Hol- 
low Circular Cast-iron Beams, and factors for use in Eqs. 91 
and 92. 




82. Transverse Strength. The transverse strength 
of Hollow Cast-Iron Beams may be ac- 
curately computed from either Eq. 89 
or 90, also from Eq. 91 or 92, and with 
much less labor. 

Example 20. — Required the centre 
breaking load of a Hollow Circular Cast- 
iron Beam, when 

. . . = 3". 875, G = 84000 pounds, 
. . . = 3M25, r, = 14000 " 
. d = 3".5, s, = 6 feet, 
" thickness of metal t = 0^375,/e, =: 1.1008 from the Table. 

By using these values in Eq. 91 and for r = 1.75, one half 
of the mean diameter the load, Z, becomes 

^ ^ 4 X 0.375 (1.75)-L1008_ X 84000 ^ ,g,, 
12 X 6 ^ 

The actual breaking load was 5122 pounds from the series of 
experiments described in Example 10. 



The outer diameter 
" inner " 
"" mean " 



CHAPTEE V. 

WROUGHT-IRON AND STEEL BEAMS. 
Section I. — General Conditions — Wrought-Iron. 

83. Coinj)ressive Streiig'tli. Crushing. — Wrouglit- 
iron, when subjected to pressure, increases its area so rapidly 
that it is impossible to determine with precision its crushing 
f^trength or that intensity of pressure that corresponds to the 
tenacity. Its value, per square inch, as quoted by the various 
writers, varies from 30000 to 90000 pounds per square inch. 

The great pressure to which wrought-iron is subjected when 
rolled into sheets and eye-beams causes it to lose a portion of 
its ductility and increase its crushing strength. The crushing 
strength of Swedish bar-iron, computed from Mr. Kirkaldy's 
experiments (Example 21), is 60000 pounds per square inch, 
while the crushing strength of wrought-iron when rolled into 
eye-beams, computed from Mr. Fairbairn's experiments, is 
86466 pounds per square inch (Example 22). 

Elastic Limit. — The compressive elastic limit of wrought- 
iron in bars may be taken at 30000 pounds per square inch. 
The computed value of the compressive elastic limit for rolled 
eye-beams is 63170 pounds (Example 31), from data furnished 
by the Phoenix Iron Company's experiments. 

84. Tensile Strength. Tenacity. — When wrought-iron 
is subjected to a tensile strain before rupture takes place, it in- 
creases in length from 15 to 20 per cent, and contracts in area 
at the fractured section about 25 per cent. Its tenacity ranges 
in value from 50000 to 65000 pounds per square inch ; wrought- 
iron diminishes in tensile strength when rolled into sheets. 



88 STRENGTH OF BEAMS AND COLUMNS. 

Elastic Lhnit. — The tensile elastic limit of bar-iron is about 
one half of its tenacity ; the mean may be considered to be 
30000 pounds per square inch. 

Steel. 

85. Compressive Strength. Crushing. — In deter- 
mining its crushing strength, the same difficulty mentioned in 
determining that of wrought-iron is encountered ; its value, as 
given for the different kinds, varies from 100000 to 840000 
pounds per square inch. 

The compressive elastic limit of steel ranges from 20000 to 
60000 pounds per square inch. 

86. Tensile Strengtli. The tenacity of steel is greater 
than that of any known material ; it ranges in value from 
60000 to 160000 pounds per square inch. 

87. To Compute tlie Compressive Strengtli. 

Formula 31, page 40, and Eq. 79, page 58, may be used to great 
advantage to determine the crushing strength for such materials 
as wrought-iron and steel whose resistance to crushing cannot 
be readily determined in the usual manner, because they in- 
crease their area so rapidly, under direct pressure, when ap- 
plied to small test specimens, that no precise determination of 
their strength can be made. The deflection of the beam, how- 
ever, must be sufficient to allow the neutral line to move to the 
position required for rupture of the fibres by the compressive 
and tensile strains at the same instant, for the reason given in 
Art. 24. 

Example 21. — Required the Crushing Yalue of C for 
wrought-iron, from the tensile strength and the centre trans- 
verse hreal'ing-clown load, when 

The depth ^ = 2.0 inches, T = 42133 pounds, by test, 

" breadth... J = 2.0 " Z = 13338 " " '' 
" span 6^ = 25.0 " m — 4. 



GENERAL CONDITIONS. 89 

The required position of the neutral line from Eq. 30 is 

3X2 



^"~2 ' 4X4X2X 



./ 9 X 4 X 2 (2)^ 42133 - 1 2 X 13338 X 25 _ 
"^ 4 X 4 X 2 X 42133" ~ ^-^^^^ 

. • . < = 2 - 0.554 = 1.446 inches and C from Eq. 31. 



^^3 X 13338 X 25 ^ ^^^^^ pounds. 
4X2(1.446)'"' ^ 

This example is taken from Mr. Kirkaldj's experiments.* 
By direct pressure he determined C = 84890 pounds, when 
the length = 2 diameters, and C= 148840 pounds, when the 
length = 1 diameter of the test specimen. Our computation of 
the value of O is made under the hypothesis that the tensile 
strain in the experiment w^as sufficiently mtense^ though it did 
not actually fracture the beam. 

Example 22. — Required the Crushing Value of C for rolled 
wrought-iron from the tensile strength and the centre trans- 
verse breaking load of a rolled wrought-iron T beam. 

The deptli of the beam d = 3'^0, T = 57600 pounds mean 
for British bar-iron. 

The breadth of the web h = 0'\5, L = 2690 pounds, from 
Mr. Fairbairn's test.f 

The breadth of the compressed flanges h^ = 2 '^5, 7)i = 4. 
The depth " " " d, = 0''.375, ^ = 10 ft. 

The position of the neutral line, the beams having no bottom 
flanges, from Eq. 30, becomes 



di = 



^ 3x3 _ / 9 X 4 X 0.5 (3f 57600 - 1 2 X 2690 X 120 ^ ^ qq ^^^ 
2 I 4 X 4 X 0.5 X 57600 



* Barlow's " Strength of Materials," p. 256. 
t Box's " Strength of Materials," p. 214. 



90 STRENGTH OF BEAMS AND COLUMNS. 

and from Eq. 37, 

0.5 (1.94)3 -f- 2.5 (0.3 75)^ (3 X 19 4 - 2 X 0-375) 
Z == 2690 = 3^<T94 >ri'2~>ri0 = ^'^^^^^ ^' 

■ •' ' ^ — 003111 ^ S^"^^^ pounds. 

Example 23. — Required the Crushing Strength of certain 
steel made by the Otis Iron and Steel Co., of Cleveland, 
Ohio, from the centre breaking load of a circular beam when 

The diameter = l'M29, T = 83500 pounds mean of 10 tests, 
" span = 20'^0, L = 3860 '' one test. 

From E(j. 78 we have 

. _ _^^_^^__ ^ 1 2S49 

•''' 4: (0.5645/ 83500 



TJ 



From the Table, page 100, <i = 1.2122 for this value of / 
and from Eq. 79, 

C ==: 1.2122 X 83500 ^ 101218 pounds. 

This example is taken from the United States Government 
Report of Tests of Iron and Steel, at Watertown Arsenal, for 

1885.* 

88. To Compute the Tensile Strength. The 

tensile strength may be computed from test values of C and 
the load Z in rectangular beams by means of Eq. 33, having 
first ascertained the position of the neutral line from Eq. 32, 
and for circular beams from Eqs. 80 and 81, page 58. 

Example 24. — Required the Tensile Strength of " Burden's 
Best " wrought-iron from the centre breaking load of a cir- 
cular beam when 

* Senate Ex. Doc. No. 36— 49th Congress, 1st Session, p. C90. 



GENERAL CONDITIONS. 91 

The diameter d = 1".26, C — 64500 pounds assumed, 

" span s =. 12'^0. L ■= 6000 " by test. 

From Eq. 80 we have 

_ 6000 X 12 ^ 
J'''~\ (0.625)' 64500 ' * 

From the Table, page 100, ^ = 1 for the above value of /*„ 
from Eq. 81, 

T =. 5^^ = 64500 pounds. 

This beam was broken at the Hensselaer Polytechnic In- 
stitute in IS^ovember, 1882, as given by Professor W. H. Burr. 

89. Transverse Strength of Wrought - Iron. 

The elastic limit load is technically called the breaking load 
for WTOught-iron beams, as from its position in the section of 
the beam where fracture should take place from the fibre 
strains being greatest, it cannot undergo the change of form 
required before its fibres can \>q fractured / this breaking load 
may be computed from the tensile and compressive elastic 
Ihnit coefiicients of wrought-iron. 

The hreahing-dowji load. — While wrought-iron in beams 
cannot be broken transversely by rupturing its fibres, yet there 
is an intensity of the transverse load at which it does not 
appear to be able to offer any further resistance to the action 
of the bending load ; this load may be called the hreaMny- 
down load ; in wrought-iron beams it is about twice the 
technical breaking or elastic limit load. 

Wrought-iron when rolled into T beams may become 
sufficiently hard and unyielding from the intense pressure re- 
quired to make it fill the rolls, that it will offer sufficient 
resistance to actually fracture the fibres by the tensile strain, as 
in Example 22. 



92 



STRENGTH OF BEAMS AND COLUMNS. 



90. Transverse Streng^th of Steel. As in the 

case of wrought-iron beams the elastic hmit load of steel 
beams is technically called the breaking load. 

The hreahing-down load of steel beams is about f tlis of the 
elastic limit load. 



Section II. — Rectangular Wrought-Iron om^d Steel Beams. 

91. Neutral Line. The position of the neutral line 
may be computed from either formula 25 or 26. From Eq. 
25 the position has been computed for the different values of 
C ^ T — q^ that are required in wrought-iron and steel rect- 
angular beams, and the results tabulated below for reference. 

TaUe of Positions of the Neutral Line in Eectangidar 
Wrought-Iron and Steel Beams : 




For ratios greater than C -^ T — Z refer to the Table, page 

80. 

93. Transverse Streng^th. Either Eq. 27 or 28 may 

be used to compute the transverse strength of rectangular 
wrought-iron and steel beams, as may be most convenient. 



RECTANGULAR WROUGHT-IROI^ AND STEEL BEAMS. 93 

ExAMPr.E 25. — Required the centre breaking or elastic limit 
load of a Rectangular Wrouglit-Iron Beam, when 

The depth tZ = 3".0, C = 30000 pounds mean value, 

'' breadth l = 1".5, T = 30000 " '' 

" span s = 33".0, q — 1, and m =- 4. 

The position of the neutral line may be computed or taken 
directly from the Table for q = 1. 

d, = 0.Y805 X 3 = 2^3415, d^ = S - 2.3415 = 0,6585. 

From Eq. 28 the value of the load, Z, required becomes 

^ 4 X 1.5 X 0.6585 (2 X 3 -f 2.3415) 30000 _^^ , 

X = -r TTT^ = 9681 pounds. 

3 X 33 ^ 

Mr. Barlow tested this iron and found its tensile elasticity 
perfect with. 22400 pounds ; the transverse 
elastic limit load was, by experiment, be- 
tween 9520 and 10080 pounds. A num- 
ber of beams tested gave similar re- 
sults. 

The deflection with 10080 pounds was 
0".963inone bar and 0".624 in another; 
hence the elastic fibre strain limits were fully developed."' 

Example 26. — Required the centre breaking or elastic limit 
load of a Rectangular Wrought-Iron Beam of Swedish iron, 
when 

The depth d = 2".0, C ^ 2263T pounds by test, 

" breadth ?> = 2".0, T = 24052 " 

'' span s= 25\0, q =i 0.94, and m = 4. 

The position of the neutral line may be computed from Ecp 
25, or taken from the Table, by proportion between the posi- 



* Barlow's " Strength of Materials," p. 278. 



94 STRENGTH OF BEAMS AND COLUMNS. 

tions for q = 1 and q = 0.9, for which d^ = 1.58. With 
these values the required load, Z, becomes from Eq. 27, 

^ 4 X 2 ( 1.58/22 637 ...^ , 

X = ---^ =. 6003 pounds. 

Mr. Kirkaldy determined, experimental!}^, that the centre 
breaking load of this beam was between 6000 and 6500 
pounds, and that the elastic limit values of C and T were as 
given above. 

The position of the neutral line for the hrealii/ng-doivn 
load of tliis beam (Exariple 21) was 1".446 below the top or 
compressed side of the beam, and as the loading progressed 
it must have moved upward from its position at the elastic 
limit, 1".58, to 1".446, its position with the hreaJcing-doivn 
load. 

The deflection with 6000 pounds was 0".378, and Avitli 6500 
0''.506 ; it only required 0".46 to fully develop the elastic 
iibre strain limits.* 

Example 27. — Required the centre transverse elastic limit 
load of a Rectangular Steel Beam, when 

The 'depth ....d^ r.75, C = 48000 pounds by test, 

" breadth h = 1''.75, C = 52000 " '' 

'' span s = 25".0, q = 0.923, and m = 4. 

The position of the neutral line, computed from Eq. 25, or 
taken from the Table by proportion, is 

d, = 1".3867 and d, ^ 0.3633. 

The required load, Z, from Eq. 27, becomes 

J. 4X1.75(1.3867)^8000 _^,^ , 

L = -3^25"^ ^ pounds. 

* Barlow's " Strength of Materials," p. 256. 



RECTANGULAR WROUGHT-IRON AND STEEL BEAMS. 95 

Mr. Kirkaldy determined by experiment that tlie elastic 
limit load of each of 4 of these beams was between 8000 and 
9000 pounds, and that the values of C and T were as 
given. 

The deflection with 8000 pounds was 0".476, and it only 
required 0".3633 to develop the elastic compressive and tensile 
flbre strain limits at the same instant.^ 

Example 28. — Required the hreahing-down load of the 
beam described in Example 27, when 

C = 1.59582 pounds by test, q = 2.3, 
T = 69336 " " m = 4. 

For the position of the neutral line from Eq. 25 or from the 

Table, 

d, = V'AVl and d^ = 0^^629. 

The breaking-down load, Z, becomes from Eq. 27, 

^ 4 X 1.75(1.12iy 159582 _..^. , 

Z= ^ — 3'^ ^^ ' = 18 ab pounds. 

The mean hreaking-down load of 4 of these beams was 
16477 pounds from Mr. Kirkaldy's experiments. The upward 
movement of the neutral line is seen in these Examples, in 
Example 27, d^ — 1^^3867 and in Example 28, d^ = l'M21. 

* Barlow's " Strength of Materials," p. 253. 



96 STRENGTH OF BEAMS AND COLUMNS. 



Section III. — Doiihh? T, Rolled Eye-Beams and IIollow 
Rectangle or Box, Wrought- Iron and Steel Beams. 

93. Nevitral Line. The neutral lines in these wrought- 
iron and steel beams are so near the bottom or tension side of 
the beam that their positions cannot be computed from ap- 
proximate formulas, and the exact formula is too complicated 
to be useful in practice. Before beams of these sections are 
maimfactured, its position should be assumed and a sufficient 
area of the metal placed above and below the assumed neutral 
line to insure equilibrium from the known values of the crush- 
ing and tensile strength of the material. 

In most practical examples of these beams the neutral line 
is located within the tension flange, and its position can be 
computed from Eq. 30, page 40, when L the load and T the 
tensile strength have been determined experimentallv, or bv 
the methods given in Problems I. and II., page 43. 

94. Transverse Strength. AVhen the position of 
the neutral line is known the transverse strength may be com- 
puted from either Eq. 37 or 38, that were deduced for the 
transverse strength of the Ilodgkinson beam, giving the letters 
of the formulas the meaning deflned in Art. 44, page 47. 

The transverse strength of both the Tee and Double- 
Headed Railroad Rails may be very accurately computed from 
the above formulas, although they are partly bounded in 
outline by curved lines. The web must extend from the 
bottom to the top of the rail, and an equivalent in area right 
line section must be formed for the metal that remains in both 
the base and head of the i-ail, which will enable the formulas 
to be applied to these forms of beams. 

95. Double T heams with tension and compression 
flanges uneq^ial m area. 

Example 29. — Required the centre elastic limit load of a 
Wrought-Iron Rolled Double T Beam, when 



DOUBLE T, ROLLED EYE-BEAMS, ETC. 



97 



Depth of beam d — 8". 38, G — 36320 lbs. required for eq'brium, 

Breadth of web h - 0".325, T = 30000 pounds mean value, 

" " top flanges 6] = 3". 175, m = 4, 

Depth " " " d^ = 1".0, s = 11 feet, 

"bottom'' (?2 = 0".38, 
Breadth " !' " 5o = 3". 675. 



The neutral line will be assumed to be 
near tlie top line of the tension flanges, 
or 

^4 == 8';05 and d^ = 0.33. 

The elastic limit load, Z, becomes from 
Eq. 37, 



IfE 



_ b _ _ b. _• 



_ I) 



^ 



— -b, 



L = 



0.325 (8.05)'^ 4-2.175 (1)^ (3 X 8.05— 2X1) 
3 X 8.05 X 132 



X 4 X 36320 = 9924 lbs., 



and from Eq. 36, 

^_ 4 X 4 (3 X 8.38 + 8.05) 30000 
"" 3 X 132 



= 9924 pounds. 



The elastic limit load was between the applied loads 9493 
and 11253 pounds from Mr. William Fairbairn's experiments, 
and their deflections were 0'^46 and 0^^60 ; the deflection re- 
quired was 0''.33.* 

Example 30. — Required the centre elastic limit transverse 
load of a Rolled Wrought-Iron Double T Beam when 

Depth of T beam d = 9". 44, C = 37165 lbs. required for eq'brium, 

Breadth of web b = 0".35, T = 30000 pounds mean value, 

" top flanges &i = 2". 4, s = 10 feet. 
Depth " " " d, = l".0, m=4. 

" " bottom " d-2=Q"A4:, f?c = 9.06 assumed, 

Breadth " " " h^ = 3". 95, d^ = 0.38 '' 



*Fairbairn, " On Cast and Wrought-Iron," p. 102. 



98 STRENGTH OF BEAMS AND COLUMNS. 

With these values the load, Z, becomes from Eq. 28, 

J _ 4 X 4.3 X 0.38 (2 X 9.4 4 + 9.06) 30000 _ ., koi o ik 
X _ 3 X 12 X 10 - ^^^^^ ^^^•' 

and from Eq. 38, 

^ 0.35 (9.06f + 2.4 (If (8 X 9.06 - 2 X 1) 



3 X 9.06 X 12 X 10 



X 4 X 37165 = 15218 lbs. 



The elastic limit load was between the applied loads 14693 
and 16373 pounds from Mr. William Fairbairn's experiments, 
and the deflection was 0''.35 and 0'^45 with these loads re- 
spectively.* 

In Examples 29 and 30 the tensile elastic limit has been 
assumed to be 30000 pounds, the mean value for wrought- 
iron, and from it the position of the neutral line has been 
computed from Eq. 30, and then the required value of C to 

produce equilibrium. 



rc 

A. 



1 



1 



\ 



rxT 



96. Double T beams with tension 

and compression flanges equal in area. 

Example 31. — Required the centre 
transverse elastic limit load of a Rolled 
Wrousrht-Iron Evebeam when 



Depth of beam d = 9".0, (J = 63170 lbs. required for eq'brium. 

Breadth of web b = 0".6, T = 30000 " mean value, 

" top flanges bi = 4". 775, m = 4, 
Depth " " " di = l".0, 5 =14 feet, 

" bottom'' d.2 = 1".0, do = 8".2, 
Breadth " " " b-i = 4".775, d.^ = 0".8. 

With these values the load, Z, becomes from Eq. 37, 



r - 0> 6(8.2f + 4. 775(1)^3X8.2-2 X 1) ^ , ^ .o... 
^' - 3 X 8.2 X 12 X 14 ^^^ ^^^'^ 



26823 lbs., 



«*, 



* Fairbairn, "^On Cast and Wrought-Iron," p 103. 



^ 



, DOUBLE T, ROLLED EYE-BEAMS, ETC. 99 

and from Eq. 28, in which ^ = 5 -j- ^2 = 5.375, 

_ 5.375x0.8(2x9 + 8.2) ^^^^^ ^ ^^^^3 

^ - 3 X 12 X U 

The vahie of T is taken at the mean, from which d^ must 
be C.8 and 6^= 63170, that equilibrium shall exist between 
the moments of the tensile and compressive resistances. 

The rolled eyebeam, of which the data given in Example 
31 is the equivalent right line section, was tested by the Phoe- 
nix Iron Co. of Pennsylvania,* and its centre elastic limit load 
was found to be between the applied loads 26880 pounds and 
28000 pounds ; the deflection was 0".572 and 0''.600 with 
these loads respectively. 

The proportion of the load and area of the section that is 
sustained by each member is given in the following table : 

Load. Area. 

Compression Flanges 12.3 31.9 per cent. 

Web 37.7 32.9 " 

Tension " 0.55 3.2 '' 

'' Flanges 49.45 24.1 " 

" " (practically lost) 7.9 " 

100.00 100.0 

The transverse elastic liimt load of the rectangular wrought- 

iron beam 5".375 X 9".0, from which the above described 

eyeheam may be supposed to have been cut, is from Eq. 27, 

when C = 30000 and T — 30000 pounds, the mean values 

for bar-iron, 

^ 4 (7.02)'' 30000 ^^.^- , 

^ ^ 3 X 12 X 14 ^ P""'"' 

From which it will be observed that the eyebeam, while con- 
taining only 31 per cent of the area of the rectangular beam, 

* Phoenix Iron Company's " Handbook of Useful Information." 



100 



STRENGTH OF BEAMS AND 'COLUMNS. 



is able to sustain 42.5 per cent of its load, which is supposed 
to be due to the elevatioji of the elastic limit during the pro- 
cess of rolling, or the top of the beam must have been laid 
with steel. 



Section \Y.— Circular Wrought-Iron and Steel Beams. 

97. Neutral I^iiie. The position of the neutral line 
in Circular Wrought-Iron and Steel Beams for the different 
ratios of C -^ T — q required has been computed, also the 
factors /e and/, for use in Eqs. 76 and 77, and tabulated be- 
low^ for reference. 







Factors for Computing the Moment of 


Eatio of Crushing 


Depth of Neutral 

Line Below the 

Crushed Side of the 


Resistance, Radius = 1. 


to Tenacity, or 






C -4- T — q. 


Beam, or dc. 


/c for Crushing 


T for Tensile 






Strain C. 


Strain T. 


3.0 


0.5375 d 


0.6468 


1.9399 


2.875 


0.5438 d 


0.6642 


1.9093 


3.75 


0.5505 d 


0.6827 


1.8766 


2.625 


0.5574 d 


0.7024 


1.8434 


2.5 


0.5646 d 


0.7230 


1.8084 


3.375 


0.5724 d 


0.7458 


1.7718 


2.25 


0.5804 d 


0.7696 


1.7314 


2.125 


0.5890 d 


0.7952 


1.6896 


2.0 


0.5980 d 


0.8228 


1.6454 


1.875 


0.6076 d 


0.8526 


1.5984 


1.75 


0.6178 d 


0.8848 


1.5482 


1.625 


0.6277 d 


0.9198 


1.4944 


1.5 


0.6404 d 


0.9580 


1.4368 


1.375 


0.6525 d 


1.0000 


1.3748 


1.25 


0.6666 d 


1.0462 


1.3074 


1.125 


0.6816 d 


1.0976 


1.2328 


1.0 


0.6976 d 


1.1548 


1.1548 



From this Table the position of the neutral line in any 
wrought-iron or steel beam may be obtained by multiplying 
the diameter d, expressed in inches, by the decimal factor 




CIRCULAR WROUGHT-IRON AND STEEL BEAMS. 101 

corresponding to the ratio C -^ T =^ q. For ratios intermedi- 
ate in value the position may be obtained by proportion. 

98. Transverse Strength. This may be computed 
from either Eq. 71 or 75, but with much 
less labor from Eq. 76 or 77, using the 
values of f^ and f^ given in the above 
Table. 

Example 32. — Required the centre 
breaking or elastic limit load of a Circu- 
lar Steel Beam when 

The diameter d = l'M29, C = 41160 lbs. mean of four tests, 

" span s =z 20'^0, T = 39200 " " '' ten " 

" f actor ... ?y2. — 1, q ^ 1.05. 

The position of the neutral line and factor, f^, becomes, by 
proportion from the Table, 

d, = 0.6912 X l'M29 = 0.780, f, = 1.132. 
With these values the load, Z, required becomes from Eq. 76, 

^ 4 (0.5645y 1.182 X 41160 _,^^^ , 

J^ = -^ — ^ = 16^6 pounds. 

From the United States Government Report of the Tests 
of Iron and Steel for the year 1885,^ the elastic limit load of 
two of these beams that were tested was 1638 pounds witli 
tlie values of O and T as given in the Example. 

Example 33. — Required the centre elastic limit load of a 
cylindrical Phoenix pin supported at both ends when 

* Senate Ex. Doc. No. 36-49tli Congress, 1st Session, p. 690. 



102 STRENGTH OF BEAMS AND COLUMNS. 

The diameter. ... d— 2".5, C = 30000 pounds from tests, 

" span 6' = 24".0, T ^ 30000 " 

" factor m = 4:, f^= 1.1548 from the Table. 

Then from Eq. 76 we have for the required load, 

_ 40:25)^1^54^^^^0000 ^ ^^3^^ ^^^^^^^ 
24 ^ 

The elastic limit strength of this pin was 11000 pounds. 
(Watertown Arsenal Report of Tests of Iron and Steel for 
1881.)^ The computed ultimate strength is 18795 pounds, 
with C and T == 50000 pounds, the recorded ultimate strength 
is 20000 pounds. 

In order that these wrought-iron pins should truly cross- 
Ireak, their deflection should not be less than 0.3024 d\ in the 
above examples the ultimate deflection was 1''.278 ; while that 
required for true cross-breaking was 0''.75, the recorded de- 
flection with 18000 pounds was 0''.78. 

A number of these pins, made of Phoenix and Pencoyd 
iron, were tested with diameters ranging from 2^ to 5 inches, 
but the span was so short, in nearly all cases, that the tensile 
flbre strain was not fully developed, causing the transverse 
elastic limit load to be not well defined and the ultimate load 
to be larger than it should be. Though the computed elastic limit 
loads do not differ very greatly from those determined by 
tests, there is a very great difference between the computed 
and experimental ultimate loads, the latter being the greater. 

This series of tests illustrates the correctness of the state- 
ment made in Art. 24, that the hend'mg moment of the ap- 
plied load at the i7icej)tio?i of the deflection of a beam is held 
in equilibrium by tlie moment of a purely compressive resist- 
ance that is distributed over the section in an uniformly vary- 



* House of Representatives Ex. Doc. No. 12, 1st Session, 47th Congress, 
p. 171. 



CIRCULAR WROIJGHT-IROIS^ AND STEEL BEAMS. 103 

iiig strain. Test E'o. Y41, page ISO, of the Eeport, was of 
Plioeiiix iron 4^' in diameter and 24'^ span. The bending 
moment of tlie centre applied load is 6 Z from Eq. 3, page 5, 
and the moment of compressive resistance is the product of 
the resultant (the area of the section by one half, (7, the elastic 
limit compressive strength), by its lever-arm, %d, the distance 
of the centre of gravity of the pressure wedge below the axis ; 
hence 

3d ,, d' € 

-8-><"4 X 2' 



^L = -;-x TT -- X ,,, 



L — — - — ^ ~ = 47124 pounds. 

384 384 ^ 



The observed elastic limit load was 48000 pounds with a 
deflection of 0^^0585, but as no correction appears to have 
been made for the settling of the beam on its bearings, we 
may conclude that the beam was on the verge of true deflec- 
tion with its transverse elastic limit load. 



104: 



STRENGTH OF BEAMS AND COLUMNS. 



Section Y.— Hollow Circular Wrought-Iron and Steel 

Seams. 

99. Neutral Line. The position of the neutral line in 
Hollow Circular Wrought-Iron and Steel Beams for the 
different ratios of C ^ T = q required has been computed, 
also the factors f^ and f^, for use in Eqs. 91 and 92, and 
tabulated below. 







Factors for Computing the Moment or 


Ratio of Crushing 


Depth of Neutral 

Line Below the 

Crushed Side of the 

Beam, or dc. 


Resistance, Rabius = 1. 


to Tenacity, or 
C+ T=q. 


/c for Crushing 


/t for Tensile 






Strain C. 


Strain T. 


3.0 


0.7500 d 


1.7320 


5.1961 


2.875 


0.7597 d 


1.7737 


5.0992 


2.75 


0.7697 d 


1.8172 


5.0000 


2.625 


0.7800 d 


1.8628 


4.8898 


2.5 


0.7906 d 


1.9105 


4.7760 


2.375 


0.8014 d 


1.9601 


4.6566 


2.25 


0.8122 d 


2.0108 


4.5334 


2.125 


0.8241 d 


2.0645 


4.3S28 


2.0 


0.8358 d 


2.1251 


4.2490 


1.875 


0.8478 d 


2.1853 


4.0966 


1.75 


0.8600 d 


2.2479 


3.9338 


1.625 


0.8727 d 


2.3135 


3.7586 


1.5 


0.8851 d 


2.3823 


3.5722 


1.375 


0.8979 d 


2.4534 


3.3728 


1.25 


0.9107 d 


2.5267 


3.1588 


1.125 


0.9233 d 


2.6024 


2.9296 


1.0 


0.9360 d 


3.6806 


2.6806 



100. Transverse Streng'tli. This may be com- 
puted from either Eq. 89, 90, 91 or 92. 

Example 34. — Required the centre breaking transverse load 
of a Hollow Wrought-Iron Cylindrical Beam, supported at 
both ends, when 

The outer diameter. . .d = 4'^0, 6^ and T = 45000 pounds, 
^' thickness of metals = 0'M875, /; = 2.6806 from Table, 
" sj3an s = 6'.0 m = 4. 



WROUGHT-IRON AND STEEL BEAMS. 105 

Then from Eq. 91 we have for the load, 

^ 4 (2y 2.6806 X 0.1875 x 45000 ^,_^ 

L = — -^ a— = 50 <b pounds. 

12 X 6 ^ 

This " tube failed suddenly with 5821 pounds." (From '' The 
Britannia and Conway Tubular Bridge," p. 135.) 



CHAPTER Yl. 

TIMBER BEAMti. 

101. The Compressive and Tensile Strengtli. 

The great number of the different varieties of timber, and 
the great variation in the strength of the timber that is 
known in different parts of the globe by the same name, 
render the determination of constants or mean values for C 
and T for use in the computation of the strength of beams a 
very difficult matter, and we find that much difference exists 
between the constants that are given for the same timber by 
the older and more recent experimenters ; especially is this the 
case for the crushing strengtli. The former make C -^ T 
less than 0.75 for all of the principal varieties, while the latter 
make it greater than unity. 

The coini)uted transverse strength of the beams broken by 
the older experimenters from their 6Y>/^6Vcr;??'.v gives, practically, 
accurate results, or such as are within the limits of the varia- 
tion in strength of the material, while those from the more 
recent experimenters make their computed transverse strength 
from twenty-five to one hundred per cent greater than their 
experiments gave, which indicates that with improved testing 
machines the more recent experimenters, such as Professor 
Thurston, Laslet and Hatfield, obtained their crushing strength 
at a point nearer the total destruction of the wood than the 
older with machines having less power. 

102. To Compute tlie Comi>ressive and Ten- 
sile Strength. Crushing. — From the fibrous charactei* 
of timber and its weak lateral adhesion it is difficult to de- 



TIMBER BEAMS. 107 

termine its crushing strength from direct pressure on small 
specimens, or that intensity of crushing strain that holds its 
tensile strength in equiUbrium : this can be accurately com- 
puted from Eqs. 31 and 79, when the breaking, transverse 
and tensile strength have been determined from experiments. 

Example 35. — Eequired the Crushing Strength of Teak- 
wood from the known tensile strength and the centre trans- 
verse breaking load of a rectangular beam when 

Depth. . .<:Z=2.0 ins., 7^=15000 lbs. mean of Mr. Barlow's tests, 
Breadth .^=2.0 " Z=938 " " " " " 

Span .... ^=7.0 feet, 7/?. =4. 

The position of the neutral line becomes from Eq. 30, 

d = ^ X ^ - , /9 X 4 X 2 (2)^ 15 000 — 12 X 938 X 12^~7 ^ q 35^^^ 
2 r 4X4X2X 15000 

. ' . d, = 2.0 - 0.35 = 1.65 inches, 

and the crushing value of O from Eq. 31, 

^ 3 X 938 X 12 X 7 ^^._ , 

6 = -. ?rm^^V2 — = IO80O pounds. 

4x2 (1.65/ ^ 

From Mr. Hodgkinson's experiments O —- 12100 pounds 
for Teak. 

Tensile /Strength. — This may be computed from Eqs. 32 
and 33, and from Eqs. 80 and 81, when the transverse and 
crushing strength have been ascertained from experiments. 

Rectangular Wooden Beams. 

103. Neutral Line. Table of positions of the neutral 
line in rectangular sections of wood, and the mean ultimate 
crushing and tensile strength of the different varieties of 
timber. 



108 



STRENGTH OF BEAMS AND COLUMNS. 



Ash, American 

" English 

Beech, American . . . 

" English 

Birch, American 

Black 

Birch, English 

Cedar, American Red 

Elm 

Fir, White Spruce . . 

" " Christiana 

Deal.... 

Oak, American Red 

" White 



English , 



Pine, Am. South-) 
em Long-leaf. . . i" 

Pine, Am. White. . . 
" Am. Yellow. . 
" Red, European 

** " Dantzic... 
" Riua .... 



Poplar 

Walnut, Black, 



Breaking Strength 

PER Square Inch 

IN Pounds. 


Eatio of 

Crushing 

to 




Tenacity, 
C 






Crushing, 
C. 


Tensile, 
T. 


f=Q. 


4400 


11000 


0.4 


5800 


14000 


0.4142 


8600 H 


12000 31 


0.7166 


5800 


15000 


0.3866 


6900 


18000 


0.3833 


7700 JI 


9000 i? 


0.4142 


9300 H 


12000 B 


0.7750 


7000 


11600 


0.6034 


4530 H 


11700 i? 


0.3872 


6400 H 


15000 31 


0.4266 


6000 


10300 


0.5825 


6830 


14000 3/ 


0.4878 


6500 31 


10000 31 


0.6500 


5850 M 


12000 J/ 


0.4875 


6000 


10000 


0.6 


7200 


18000 


0.4 


9100 


18000 


0.5055 


6500 H 


10000 31 


0.65 


9500 H 


19000 31 


0.5 


8000 


12600 


0.6349 


8000 


19200 


0.4166 


5000 


10000 


0.5 


5400 M 


12000 31 


0.*45 * 


7500 3f 


14000 M 


0.5357 


5400 31 


8000 31 


0.575 


574^ H 


11549 B 


0.4977 


6586 // 


12857 B 


0.5122 


5100 T 


7000 T 


0.73 



Depth of 
Neutral 

Line Be- 
low the 
Crushed 
Side of 

the Beam, 
or 6?c. 



Authorities. 



0.8900 dB., Barlow. 
0.8877 d H., Hodgkinson. 
0.8267 ^^M., Molesworth. 
0.8933 rZR., Rankine. 
0.8941 d 
0.8878 d 
0.8165 d 

0.8477 d 
0.8939 d 
0.8843 d 
0.8518 d 
0.8706 d 
0.8376 d 

0.8711 d 
0.8495 d 
8900 d 
0.8673 d 
0.8376 d 
0.8685 d 
0.8410 d 
0.8525 d 
0.8685 d 

0.8792 d 
0.8559 d 
0.8533 d 
0.8690 «Z 
0.8660 d 
0.8243 d 



Tlie position of the neutral lines in the above Table was 
computed from Eq. 25, d being unity. 

104. Transverse Strength. Tim- 
ber beams breaking with a well-defined 
fracture in its fibres, the transverse 
strength may therefore be computed from 
the crushing and tensile strength by 




TIMBER BEAMS. 109 

means of either Eq. 27 or 28, as the rectangular is the form 
that is principally used in wooden beams. 

Example 36. — Required the uniformly distributed breaking 
load of an American White Pine Rectangular Beam, when 

The depth d = 14'^0, C = 5000 pounds mean of tests, 

" breadth.... J = 6^0,^=10000 '' " '' 

" span s — 28'. 0, m = 8, Art. 31, and q = 0.5. 

The position of the neutral line from the Table is 

r/e = 0.8685 X H = 12.16 inches, 
and the breaking load from Eq. 27 becomes 

^ 8X6 (12.16)' 6000 _„„^ . . 

^ - 3X12X28 ' = ^'^^^ P""^^'- 

The Table in Trautwine's Engineers' Pocket-Book gives 
37800 pounds as the breaking load of this beam. 

Example 37. — Required the breaking load of a Rectangular 
English Oak Beam fixed at one end and loaded at the other, 
when 

The depth d = 2".0, = 6500 pounds mean by test, 

" breadth b = 2'^0, T = 10000 " '' " 

" span c9 = I'.O, q = 0.65 and m = 1, Art. 34. 

The position of the neutral line is, from the Table, 
d^ = 0.8376 X 2 = 1.6752 inches, 
the load, Z, from Eq. 27, becomes 

Z = lX2(m52y65O0 ^ 253 
3 X 12 X 1 ^ 



110 STRENGTH OF BEAMS AND COLUMNS. 

From Colonel Beaiifoy's experiments the mean breaking 
load of 6 of these beams was 258 pounds each.* 

Example 38. — Required the centre breaking load of the 
beam described in Example 37, when 

The span = 7.0 feet, m = 4. 

The required load, Z, from Eq. 27, becomes 

L ^ iX_2_(M752):65^a0 ^ ^^^ 
3X 12X 7 ^ 

From Mr. Barlow's experiments the mean breaking load of 
three of these beams was 637 pounds each. 

Example 39. — Required the centre breaking loads of the 
following Rectangular Wooden Beams, when. 

The depth ^ = 2".0, C = values from the Table. 

" breadth h =z 2'\0, T = " " " 

'' span 5 = 50".0, m, =^ 4:. 

The position of the neutral line is obtained by midtiplying 
the depth, >/, by the factor corresponding to the ratio C ^ T 
given in the Table, for each case. 

Computed. 2 tests. 

Christiana Deal. .Z = o T ;^n -= 969 940 and 1052 lbs. 

d X 50 

English A«h ^^ 4x2a.6534y 8600^ 1304 an<U 304 - 

*= 3 X '^O 

T^ VI -D- 1 T 4 X 2 (1.7686y^ 6400 ,„_ ,,^. a ^oc^A ^< 

English Birch. . .L — -^ — = 1067 1164 and 1304 

^ 3 X 50 

Am. Black Birch L = 4 X 2 (1 .6954)^TO()0 ^ ^^^^ ^^^^ ^^^ ^^^^ . . 

3 X 50 

These exjierimental breaking loads are taken from Mr. P. 
* Barlow's " Strength of Materials," p. 58. 



TIMBER BEAMS. Ill 

W. Barlow's experiments ;* the values for C and T are the 
mean that are usually quoted by authors. 

Example 40. — Required the centre breaking load of a Rect- 
angular French Oak Beam, when 

The depth <^ = T.5 inches, C = 6000 pounds, 

" breadth h= 7.5 " T = lOOOO '' 

" span s =^ 15.0 feet, q = 0.6 and m = 4. 

The position of the neutral line from the Table becomes 
d^ ■= 0.8495 X 7.5 = 6.374 inches, 
and the required load, Z, from Eq. 27, becomes 



Z = i>^5 (6.374)- 6000 ^ ^3,^3 
3 X 12 X 15 ^ 



M. Buff on, in experiments made for the French Govern - 
ment,f broke two of these beams with 13828 and 14634 
pounds respectively ; from Rondelet's experiments the values 
of C and T are supposed to be equal to those given in the 
Table for American Red Oak. 



Circular Wooden Beams. 

105. Neutral Line. The position of the neutral line 
and the value of the factors y^ and/!r for use in Eqs. 76 and 
77 may be obtained from the following Table, by proportion 
when necessary : 

♦Barlow's " Strength of Materials," p. 86. f Ibid., p. 56. 



112 



STKENGTH OF BEAMS AND COLUMISTS. 



Ratio of Cruehing 
to Tenacity, 


Depth of Neutral Line 

Below the Crushed Side of 

the Beam, or 


Factors for Computing the Moment 
OF Resistakce, Radius = 1. 


or 






C-i- T-q, 


da. 


/c for Crushing 
Si rain, C. 


/r for Tensile 
Strain, T. 


1.0 


0.6976 d 


1.1548 


1.1548 


0.9829 


0.7000 d 


1.1632 


1.1433 


0.9473 


0.7050 d 


1.1812 


1.1189 


0.9126 


0.7100 6^ 


1.1991 


1.0943 


0.8790 


0.7150 d 


1.2174 


1.0708 


0.8463 


0.7200 d 


1.2357 


1.0458 


0.8133 


0.7250 d 


1.2540 


1.0200 


0.7837 


0.7300 d 


1.2726 


0.9974 


0.7541 


0.7350 d 


1.2912 


0.9732 


0.7245 


0.7400 d 


1.3098 


0.9490 


0.6965 


0.7450 d 


1.3287 


0.9251 


0.6686 


0.7500 (^ 


1.3476 


0.9013 


0.6423 


0.7550 d 


1.3666 


0.8775 


0.6161 


0.7600 d 


1.3856 


0.8538 


0.5912 


0.7650 d 


1.4048 


0.8304 


0.5664 


0.7700 d 


1.4240 


0.8066 


0.5429 


0.7750^ 


1.4434 


0.7832 


0.5194 


0.7800 d 


1.4628 


0.7598 


0.4971 


0.7850 d 


1.4828 


0.7365 


0.4749 


0.7900 d 


1.5020 


0.7133 


0.4534 


0.7950^ 


1.5216 


0.6895 


0.4320 


0.8000 d 


1.5412 


0.6658 


0.4129 


0.8050 d 


1.5611 


0.6442 


0.3938 


0.8100 6? 


1.5810 


0.6226 



106. Transverse Strength, The transverse strength 

of circular wooden beams may be com- 
puted from either Eq. 74 or 75, also from 
Eq. 76 or 77, with the aid of the values of 
the factors f^ and f^. deduced from the 
above Table. 

Example 41. — Required the centre 




Deal Beam, when 



breaking load of a Circular Christiana 



The diameter d = 2".0, C ~ .5850 pounds mean of tests, 
" span ... s = 48".0, T =: 12000 " " " " 

m =4, q= 0.4875. • 



TIMBER BEAMS. 113 

The required load, Z, becomes from Eq. Y7, 

L = ill )' 0-T26 X 12 000 ^ ^^g ^^^^^^^_ 

Mr. Barlow broke tliree of tliese beams with 710, 796 and 
780 pounds respectively, the mean being 772 pounds."^ 

107. Relative Strength of Square and Circu- 
lar Timber Beams. The required relation will be ob- 
tained from either Eq. 83, 81 or 85, as the case may require. 

Example 12. — Required the centre breaking load of a Cir- 
cular Christiana Deal Beam from that of the circumscribed 
square beam of the same span and material, when 

Side of the square ... 6? = 2".0, (7= 5850 lbs. mean of tests, 
Diameterof the circle 6?= 2".0, 7"=: 12000 " " " " 

Span s = 18^0, ^ = 0.1875, 

y, from the Table, page 108, becomes 0.8711 when d = 1, 
and/*c, from the Table, page 112, becomes 1.1912 when ^ = 1. 

With these values Eq. 83 becomes 

3/e _ 3 X 1.1912 ^ ^ ^3^. 
8/ ~ 8 X (6.8711)' ' ' 
hence. Circle = Square X 0.737. 

Breaking strength of 2" sq. beam, Mr. Barlow's tests, 1117 lbs. 

" " " 2" circ. " 1117 X 0.737 823 " 

Mean breaking strength of 3 beams, Example 39 ... . 772 " 

Mr. Barlow says that the 2" square and the 2" diameter cir- 
cular beams were cut from the same plank, " which was a very 
fine specimen of Cliristiana deal." The breaking strength of 
the 2" square and 48'' span beam, with the above values of T 
and G, should have been 986 pounds, and the circular beam 
986 X 0.737 = 726 pounds, as in Example 11. 

* Barlow's " Strength of Materials," p. 78. 



CHAPTER YII. 

STRENGTH OF COLUMNS. 

108. General Conditions of Failnre of Col- 
nniiis. Enler and Tredgold are credited with the only well- 
recognized attempts that have been made to deduce rational 
formulas for the strength of columns, but the basis of each 
of their theories involves the assumption of the existence of 
conditions that render their application to the actual pheno- 
mena observed in practice inapplicable without the aid of 
empirical factors determined from experiments. 

The law^s that govern the breaking strength of pillars were 
investigated, experimentally, by Mr. Eaton Hodgkinson in 
1S40, but no rational theory has ever been advanced that will 
explain the phenomena observed by him and subsequent in- 
vestigators ; and our constmctors are to this time using the 
empirical rules of either Hodgkinson or Gordon, that were 
deduced from the former's experiments, to compute the re- 
quisite dimensions of their pillars, or as they mny have been 
modified to conform to the results obtained by subsequent in- 
vestigators from new material or that produced from improved 
methods of manufacture. 

The effect that will be produced upon a given piece of 
material, when subjected to a strain in the direction of ita 
length, depends entirely upon its deflection. The defiection 
varies with the material and with the ratio of the length to 
the least diameter. It has been found, experimentally, accord- 
ing to Mr. Tredgold, '' that when a piece of timber is com- 
pressed in the direction of its length, it yields to the force in 
a different manner according to the proportion between its 



STRENGTH OF COLUMNS. 115 

length and the area of its cross-section," which is according 
to the amount it is able to bend or deflect laterally. 

The material of a pillar when subjected to an applied load 
in the direction of its length, in order to avoid the strain thus 
brought upon its flbres, deflects laterally and assumes a form 
composed of one or more curves, as its ends may be round or 
flat ; a round end pillar assumes the one curve form and the 
flat end pillar takes a form made up of three or four curves. 
The former is represented in Fig. 34, and will be called the 
Triple Flexure form ; it is that in which the pillar offers the 
least resistance to breaking by an applied load. 

When a pillar of the Triple Flexure form fails with deflec- 
tion, tension exists in its fibres at g (Fig. 3-1), and compression 
at m ; hence there must be a point in the line gam.^ at which 
there is no strain ; likewise there must be a point without 
strain in the line gds ; and compression existing in the flbres 
at the points <9, f and Z", the entire side, kef ho, must be in 
like condition. The fibre strains at two sections of the pillar, 
such as ao and db, must therefore increase uniformly in inten- 
sity from zero at a and d, to its greatest at c {ind Z>, respec- 
tively. The failure of a few columns along the lines etc and 
dh indicates that they join the points of 'reverse curvature of 
each side of the pillar, though a fracture beginning at c 
may follow the line of less resistance, cm. There is also a 
point, n, in the line g y, at which the strain is zero in intensity, 
and a curved line, a n d, connecting these zero points is the 
neutral surface of the column. 

Since the column must bend symmetrically there is a point 
in each of the lines ac and dh that is one fourth the length of 
the column from each of its ends, nik and so, respectively, 
thus making the middle curve of the triple flexure form one 
half of the length of the column. Tlie angle that the lines 
ac and dh make with the plane of the ends varies with the 
material, but in all pillars it is supposed to approximate the 
angle of 45 degrees. 



116 



STRENGTH OF BEAMS AND COLUMNS. 



However short a pillar may be when loaded, it will attempt 
to assmne the triple flexure form. The curves above and below 
the lines ac and clh being the first to take their shape, thus 
give rise to the various phenomena that are observed when 
short blocks of granular textured material, such as cast-iron, 

are crushed. When the points c, f and h 
coincide, the block is sheared across at one 
plane ; when a coincides with m and d 
with 8 the block splits up into four or 
more wedges. 

The strength of a j^illar generally dimin- 
ishes as it becomes longer in proportion to 
its least diameter. Should a series of pil- 
lars of an uniform section be constructed 
with progressively increasing lengths and 
the strength of each be obtained by ex- 
periment, there would be found one length 
whose strength per square inch of section 
would be greatest ; the strain must be 
uniformly distributed over the section of 
the pillar and is the crushing value of C 
for the material. As the length increases 
the strength decreases, until a length of 
pillar is reached whose strength is just 
one half of the greatest strength above 
described ; the strain must be vmformly 
varying in intensity — zero at g and great- 
Fxg. J4 est at f^ where its intensity per square 

inch is the crushing value of the material. The pillar of our 
series, with this ratio of length to its least diameter, will fail 
by crushing without deflection when it is on the verge of fail- 
ing by crushing with deflection ; twice the mean intensity of 
its strength per square inch of section is the crushing strength 
of the material, or that criishing strength that equilibrates 
the tenacity when beams and columns fail by cross-hreahing . 




STRENGTH OF COLUMNS. 117 

At the inception of the deflection of the pillars of our series 
that fail with deflection, the neutral line coincides" with the 
side, 7ngs^ of the pillar ; as its length continues to increase it 
will now be able to deflect, all compression having been re- 
moved from one side, and the neutral line will move from <j 
toward f, a distance equal to the deflection, until the deflec- 
tion, gn^ equals that depth of extended area that was found to 
be necessary for equilibrium when a beam of the same material 
and section is broken by a transverse load, and tlie pillar will 
fail with all of the phenomena of true cross-breaMng in a 
beam. 

From the length at which a pillar of our series just begins 
to deflect, to that at which it deflects sufficiently to cross-hreak^ 
the anomaly of the strength of pillars, increasing with an in- 
crease of length, is exhibited ; of two pillars of the same section 
and material, the one that deflects most, if it breaks within 
the limits of deflection above described, will sustain the 
greater load. When the length is increased beyond that at 
which the pillar first breaks by true cross-breaking, its strength 
again diminishes, and the further anomaly is j^resented of a 
column failing by cross-breaking with the same load that a 
column of the same section and shorter length will fail with 
by crushing, but with less deflection. 

The pillar by deflecting seeks to relieve itself of the resting 
of the load directly on its middle section, but this it can only 
do to the extent of gn = d^, or that value required for rupture 
in beams, and it must fail by the direct pressure of the load 
along the line 7?/, as it is being transmitted to the foundation, 
so, for all lengths of 2:)illars ; but when it deflects so much that 
the resultant of the applied load passes without the middle 
section of the pillar, it thus increases its effect by a cantilever 
strain, which effect must be deducted from the amount of the 
tivo pressure wedges that are required along nf to rupture the 
fibres by both tension and compression, as explained in Art. 
109. 



118 STEEI^GTH OF BEAMS AND COLUMNS. 

Pin End Columns. 

The " failure " of long pin end columns or those that fail 
l)j cross-hreakmg, presents some interesting peculiarities. The 
frictional resistance offered to the movement of the pillar 
around the pin, as it deflects, increases with the size of the 
pin ; it varies in pillars of like dimensions and material when 
tested with pins of the same size, and decreases, in all cases, 
after the pillar has been fairly set in motion by the lateral de- 
flection. Thus, the results of experiments sIioav that a load 
less than that required to produce fracture by compression 
will be sustained, but as the deflection approaches near to r/^ 
in value, the amount of friction becomes so small that the pil- 
lar will " suddenly spring " to a deflection that will cause the 
pillar to fail with this load by true cross-breaking, and of two 
like columns, one will sustain for a time a greater load than 
the other, on account of greater frictional resistance around 
the sofiit of the pin, evidenced by its deflection being the least ; 
the greater load will then commence to decrease, with a freer 
motion of the pillar, and will finally reach a point wliere it 
will " suddenly spring'- to a cross-hrealting deflection, thus 
breakino; with a less load than it had alreadv sustained, but with 
less deflection. That the pillar is truly broken by its " sudden 
spring " is evidenced by the fact that upon being released from 
the pressure of the load it will now require with wrought-iron 
pillars, in many cases, less than one half of the original load 
to produce the greatest deflection before obtained. This un- 
certain and variable frictional resistance around the pins of 
pin end columns gives rise to many anomalous results that 
would not otherwise be encountered in making a series of 
tests with pin end pillars when made of homogeneous material, 
such as wrought-iron, and renders it impossible to determine 
the exact law that s^overns their failure without eliminatins: its 
irregular effect. The momentum of the sudden spring fre- 
quently causes the observed deflection to greatly exceed the 



STRENGTH OF COLUMNS. 119 

true deflection at failure, and tlie load sustained at the instant 
of springing often exceeds the true cross-breaking load. These 
sources of erroneous observation become less in effect as the 
size of the ])\i\ is increased and as the pillar becomes shorter 
in length, and to increase in effect with an increase in tlie 
length of the pillar and a decrease in the size of the pin. 

109, Kesistaiice of the Cross-Section of the 
Column. From the preceding Art. it is apparent that when 
a pillar fails by crushing without deflection, its strength is 
simply the product of its area by the mean intensity of the 
pressure per square inch, which varies with its length, and that 
after it begins to deflect, equilibrium nmst be established be- 
tween the inoinents of the applied and resisting forces with 
reference to \\\q fulcrum or origin of monients,y (Fig. 34), the 
neutral line being at n^ the line of direction of the forces, 
perpendicular to the section fg^ and uniformly varying in in- 
tensity. 

When a column fails with a deflection that compels the re- 
sultant of the applied load to pass through the middle section of 
the pillar, it will only be necessary to find the amount of direct 
pressure along the line /^/that will cause the pillar to " fail." The 
tensile strain along the line gn must be held in equilibrium by 
an uniforirily varying crushing strain along the line 7if\ the 
maximum intensity of this pressure wedge is always less than 
the crushing value of C for tlie material, until gn = d^^ which 
is required for rupture in a beam, when it becomes equal to 
it. It will also require another pressure ivedge along this 
same line nf to crush the material, independently of the^r^,s'- 
sure iredge that is held in equilibrium l)y the tensile strain 
along the line gn ; hence the actual pressure required along the 
line nf is the sum of the two 2)ressure wedges, one that crushes 
the material direct, and the other that ruptures the exteiided 
portion of the pillar ng. When gn = d^^ these two press^ire 
wedges become equal in size, and they must be constant in value 



120 STRENGTH OF BEAMS AND COLUMNS. 

for all lengths of pillars that fail bj true cross-hrealcing^ and 
their sum is the breaking load of the pillar as long as the re- 
sultant of the applied load passes through the middle section 
of tlie pillar. 

The uniformly distributed load on the top of the pillar at 
mh, by the deflection, has been converted into an uniformly 
varying load at the section fg ; it will so continue when the 
pillar becomes a cantilever, being zero in intensity next the 
concave side of the pillar and greatest at the tangent line con- 
necting the points h and <9, and its moment or power to break 
the pillar as a cantilever must be computed with reference to 
the fulcrum y, its lever-arm will be {p — \(T) for the rectangle, 
[d — f c/) for the solid circle, and (p — \(l) for hollow circles, 
in which d is the least diameter and d the deflection. The 
direct tensile and compressive strain along the line gf that 
is recpiired to hold in equilibrium the inoment of the applied 
load acting upon the pillar as a cantilever must be deducted 
from the sum of the two pressure wedges that are required 
to cross-break the pillar, in order to obtain its true breaking 
load. 

Columns can be constructed of wrought-iron, steel, w^ood, 
and some cast-iron that will '' fail " by crosH-hreahing^ with 
the full value of the sum of these two equal pressure wedges ; 
but with most cast-iron and other material in which C -^ 7^ is 
approximately greater than 1.75 for solid circles, 3.00 for hol- 
low circles and 2.0 for rectangles, the pillar becomes a canti- 
lever before it deflects sufficiently to cross-break. 

The sum of these equal pressure wedges is the greatest load 
that a pillar can bear at the instant of " failure by cross-break- 
ing," but experimenters frequently obtain for ^liort pillars a 
greater apparent breaking load, from the fact that as the re- 
sultant of the load passes through its middle section, the pillar 
cannot at tlie instant of iYUQ/aihore escape from under it, and 
new conditions are set up, such as making two pillars, etc. 

The formulas deduced for failure of pillars by cross-hreaJc- 



STRENGTH OF COLUMNS. 121 

ing from this theory are identical in form with those giv'^en by 
Mr. Lewis Gordon, and it can thus be seen why, that although 
his formulas are deduced from impossible conditions of cross- 
breaking, they can be so modified by the teachings of experi- 
ment that very accurate results may be obtained from their 
use. 

110, Notation. The following notation has been 
adopted from previous chapters, and will have the same 
significance wherever it is used. Other special notation will 
be given as it may be required. 

C — the greatest intensity of the compressive strain in lbs. per square inch. 
T = the " " " tensile 

q = the quotient arising from dividing C by T. 

b = the width of the section in inches. 

d = the depth in the direction in which flexure takes place in inches, 
^c = the depth of the compressed area in inches. 
di = the " " tension " " 

L = the breaking load in pounds. 

I = the length of the pillar in inches. 

d = the deflection in inches. 

111. Deflection of Colnnins. In the present state 
of our knowledge of the cause of flexure, the laws that deter- 
mine the amount that a pillar will deflect at the instant of 
failure can only be obtained from a carefully conducted series 
of experiments. On account of the expense and the great 
pressure required, the number of experiments on the breaking 
strength of pillars of the size used in structures is very small 
and incomplete for all lengths required for a complete knowl- 
edge of the subject ; and as the theorists have taught, following 
Euler, that the strength of a column is independent of its 
deflection, it has caused the experimenters to so regard it and 
pay it but little attention, their means of measuring it being 
crude, and the record made in an unsatisfactory manner. 

The amount of deflection that is required, before a column 



122 STRENGTH OF BEAMS AND" COLUMNS. 

can be broken, is, comparatively, a very small qnantitj, mncli 
less than that required for rupture should the pillar be broken 
as a beam, and requires very accurate means for its measure- 
ment. The amount that a pillar deflects is usually stated to 
be the distance that its middle section " moves," measured 
from a given stationary point, which is only correct when the 
points Ixyf and o (Fig. S^t) lie in the same right line at the 
beginning of the movement ; should f lie to the right of the 
line ko^ the observed deflection will be too great by its distance 
from the line, and when /' lies to the left of the same line Z'6>, 
the observed deflection will be too small by its distance from 
the. line Tio^ which must be parallel to the line of direction of 
the resultant of the applied load. The deflection is an inci- 
dental quantity ; the distance required is the lever-arm of the 
resultant of the load, which is the perpendicular distance from 
the fulcrum, y, to its line of direction. 

I z=z the length of the pillar in feet and decimals, 
d^ =r the depth of the compressed area in inches, 
c = a constant quantity determined from experiments, 
S = the breaking deflection in inches. 

Then 

'-'i (93) 

The factor, c, must be determined for the different material 
used in structures, from experiments on fixed, round, two pin, 
and on one pin and one fixed end columns ; it has a different 
value for each material and style of end connections, and 
probably for each different shape of cross-section. 

Experiments. 

The experiments required to determine the laws that 
govern the deflection of colunms of a given section and 
material may l)e very much simplified by determining defi- 
nitely the following points : 



STRENGTH OF COLU.AINS. 123 

1st. The ratio of I ^ d that gives the true crushing vahie 
of the materiah 

2d. The largest ratio oi I -^ d that sustains its greatest 
load without deflection. 

3d. The smallest ratio of I ^ d that eross-hreaks. 

4th. The largest ratio of / -^ d tliat cross-hreaks without 
leverage of the appKed load. 

The law governing the change in the strength from the 1st 
to the 2d and from the 2d to the 3d ratio would then be re- 
quired ; from the 3d to the 4th ratio the strength would be 
constant, and for larger ratios than the 4th the deflection, or 
lever-arm of the load, would be the only varying element, and 
only for these pillars will it be necessary to determine the 
values of the factor c, in Eq. 93, for the various kinds of 
material used in structures. 

JSTo attempt will be made to determine, for any given 
material, the limits above defined, from the incomplete record 
of experiments at present existing. 

112. Classification of Pillars. From the descrip- 
tion of the manner of " failure " of columns of various 
lengths, it is evident that they may be divided into the follow- 
ing general classes : 

1st. Pillars tHxVt fail by crushing. 

2d. Pillars that fail by cross-breaking. 

These classes may be again divided into Cases, which, for 
convenience of reference, will be numbered consecutively. 

Pillars that fail hy crushing. 
Case I. — Pillars that fail with the full crushing 

STRENGTH OF THE MATERIAL. 

Case II. — Pillars that fail with less than the full 

CRUSHING strength OF THE MATERIAL AND WITHOUT DEFLEC- 
TION. 



124 STRENGTH OF BEAMS AND COLUMNS. 

Case III. — Pillars that fail with less than the 

FULL CRUSHING STRENGTH OF THE MATERIAL AND WITH DE- 
FLECTION. 

Pillars that fail hy cross-hreaking. 
Case IV. — Pillars that cross-break from compression. 
Case V. — Pillars that cross-break from compression 

AND CANTILEVERAGE. 

End Connections. 

Columns are again classed from the form or shape of their 
end connections, while thej all fall under one or the other of 
the iive cases given above. In columns of a given material and 
dimensions, the only element of variation in their strength aris- 
ing from differences in its end connections is the deflecticm. 

A flat or fixed end column is one whose ends are planes 
at right angles to its length ; for a given material and dimen- 
sions it deflects less, and is, therefore, the form of greatest 
strength. 

A round end column is one whose ends are spherical ; con- 
sequently, being free to move laterally, it deflects most and is 
the form of least strength. 

Pin end is a class of columns used in bridge construction 
in which the load is applied to pins passing through holes in 
its head and foot, and by them transmitted to the column. 
The smaller the pin and the less frictional resistance that may 
be developed between the ^nn and its soffit^ the nearer it ap- 
proaches the condition of a true round end column, and the 
less strength it will exhibit, while the larger the pin and the 
greater frictional resistance that may be developed tlie nearer 
it approaches the condition of a flat end pillar, and the greater 
strength it will exhibit. 

One pin and one flat end^ in its strength, follows laws 
similar to the above and a mean between flat and round end 
pillars. 



STRENGTH OF COLUMT^S. 125 

Mateeial and Section. 

In practice, columns are again classed as wooden, cast-iron, 
wrouglit-iron and steel, from the material of which thej are 
made, and from the shape of their cross-sections into rect- 
angnlar, circular, hollow circular, angle-iron, box, channel, eje- 
beam, tee, etc. From the incomjDlete tests at present made 
the lengths of columns of a given section and material that 
belong to each of the five cases of failure given above cannot 
be determined. In the examples quoted in the sequel these 
limits are, however, well defined in some special cases of 
material. 

113. Case I. — Columns that fail with the full crush- 
INO strength of the material. 

Let A represent the area of the section of the pillar in 
square inches, then as the load must be umforinly distributed 
for this case, the crushing load will be 

L^A-xC. ■ (94) 

114." Case II. — Columns that fail with less than the 

FULL CRUSHING STRENGTH OF THE MATERIAL AND WITHOUT DE- 
FLECTION. 

In cdl pillars of this class, at failure, the strain at the surface 
of one side is constant and equal to the erusliing value of C 
for the material ; the strain at the surface of the opposite side 
varies with the length, from the crushing value of C to zero in 
intensity, at which length the strain is uniformly varying, the 
mean intensity being 6^ -^ 2. 

l^ = the length of pillar that fails with full crushing strength, 
I = the " " '^ " " " a mean intensity, (7^2, 
X = the " " " whose strength is required, 
A = the area of the section of the pillar in square inches. 

The values of l^ and I vary with the material and are de- 
termined experimentally ; assuming that the intensity of the 



12Q STRENGTH OF BEAMS AND COLUMNS. 

strain on one side of the pillar varies with the length, then for 
the length x it will he C ( -, V I and the mean for the sec- 



(^3' 



tion will be (^+^(7 v )) "^ 2? and the load that the 

pillar of the length x will fail with is 

For the longest column to which this formula applies, Z = a?, 

and Eq. 95 becomes 

A y C 
L = ^t (95A) 

To Compute the Crushing Str'mgth. — The above described 
manner of failure of pillars and the formulas resulting may 
be used very advantageously to compute the crushing strength 
of any material, which, in this case, is not affected by the 
tensile strength. And we are thus furnished w^th a valuable 
means of testing the correctness of the computed crushing 
strength for the same material when broken in a beam by the 
methods given in Articles 39 and 55 ; also that obtained from 
direct experiments with short blocks. 

Deducing the value of C from Eq. 95A, we have 

9 V 7^ 

C = ^-AA, . (^^5B) 

which is the formula required from which to compute the 
crushing strength of the material in a pillar that fails or sus- 
tains its greatest load 'without deflection wdien it is on the 
verge of failing with deflection. 

Example 43. — Required the greatest strength of a Rect- 
angular Wrought-lron Column, tested with two l"h pin ends, 
when the deflection (^ = 0, area A — 8.85 square inches, and 



STRENGTH OF COLUMNS. 127 

— 50000, assumed to be equal to the tensile strength ob- 
tained from direct tests. 
From Eq. 95 A we have 

^ ^ aS^X^OC) ^ 231250 pounds. 

From the United States Government Watertown Arsenal 
Tests, 1882-1883,^ the greatest strength of nine of these col- 
umns was found by experiments, the mean was 234850 pounds, 
the deflection varied from 0''.02 to 0'M2, the mean was0".07; 
the lengths varied from 5-1: to 78 inches, and they were all 
approximately three inches square. The mean strength of 
four of these columns, tested with one flat and one \"\ pin 
end was 210800 pounds, the mean deflection, 0".12 ; the 
length of two was 90 inches and that of the other two columns, 
120 inches. 

The mean strength of four of these columns, tested ^\\\\flat 
ends, was 217875 pounds, the mean deflection was 0''.13, the 
lengths were 90 and 120 inches. 

115. Case III. — Columns that fail with less than the 

FULL CRUSHING STRENGTH OF THE MATERIAL AND WITH DEFLEC- 
TION. 

The pillars of this class vary in length for any material and 
section, between the limits of I of Case XL, or that length at 
which the mean intensity of the crushing load is 6^ -^ 2, and 
that length of Case lY. that flrst fails by cross-breaking, both 
limits being determined by experiment. 

T' = the greatest tensile strain at the middle section. 

C = the " compressive strain required to balance T\ 

Then assuming that the intensity of the tensile strain in the 
convex side of the pillar varies with the deflection, ^, from 

* Senate Ex. Doc. No. 5— 48th Congress, 1st Session, pp. 60-67. 



128 STRENGTH OF BEAMS AND COLUMNS. 

to T^ the tenacity of tlie material, the following formulas are 
deduced for the various sections : 

.'. C = q- -J-. 

By giving to q and ^ the proper value for the section, the 
following formulas have been deduced : 

Rectangular Pillars. 

From Eqs. 27 and 28, page 38, we have 




^ 



_ T K -f- — cj ^^ which cL = ^, 



d, 



Then the expression for C ' becomes, from 
substituting these values, 

^, ^ S' iZd - 6) T 
d, {d - df ' 

The sum of the greatest intensities of the two pressure 
wedges at '^ failure" is O -\- C . 

The value of d^. in this formula is that required for rupture 
of the column when broken as a beam. 

Example 44. — Required the greatest strength of a Rect- 
angular AVrought-Iron Column, tested with two 1''^ pin 
ends, when 

The diameter ^ = S'^OO, C = 50000 pounds assumed, 

" breadth h = 3".00, T = 50000 '' by test, 

'' deflection d = 0".34, d^ = 0''.66 for rupture. 



STRENGTH OF COLUMNS. 



129 



From Eq. 96 we have 

T^-f^nnnn i (0-34)2 (3x3 - 34) .^,^„„~| 3 (3 - 0.34) ^,-^.„ , 

L —\ 50000 A TTTTT-r. 7rn-.-T- -^0000 = 2422d8 pounds. 

L O.bb 1,3 — 0.34)- J 2 

Tlie greatest strengtli was 284000 pounds,* tlie length was 
80 inches. 

Circular Pillars. 

Bj substituting C -\-C' for 2(7 in Eq. 100, page 131, whicli 
gives the resultant of the load for this case, we have for the 
load that the pillar will fail with, 



Z = 



[('+?f].-/ 



(97) 



The numerical values required for the 
factors q and f are to be taken from the 
Table, page 131, for that position of 
the neutral line that corresponds to 
d^ = d —6. The value of d^. is that re- 
quired for rupture of the material in a 
circular beam, with its full compressive 
and tensile strength. 

Example 45. — Required the greatest load sustained by a 
Cylindrical Column of Midvale Steel, when 




The diameter d = 1^M29, C = 
" length.... Z = 8".96, T = 
" deflection S = 0''.25, o = 



152000 pounds mean of tests, 
112285 " " " " 

1.353. 



From the Table, page 131, y = 1.1 839, for the neutral line 
at failure, d, = l'M29 - 0.25, q = 0.5278, and d^ = 0.38, 
the position of the neutral line of rupttire. 



* From the United States Government Watertown Arsenal Teste, 1883, 
Senate Ex, Doc. No. 5— 48tb Congress, 1st Session, page 56. 



130 STRENGTJI OF BEAMS AND COLUMNS. 

From Eq. 97 we have 

L = (152000 + ^ ^5 X Q-^J^^^^ X ^^^^^'^ -^ (0.5645)-^ 1.1839 = 72056 pounds. 
The greatest strength was 81250 pounds."^ 

Hollow Cikculak Pillars. 

By siibstitutnig O -\- C for 2(7 in Eq. 101, page 135, which 
gives the resultant of the load for this case, w^e have 

Z=[C+2JI ]/■//„ (98) 

in which the numerical values required for q and ^/^ ^re to be 

taken from the Table, page 136, for that 

~P^^^^^^ position of the neutral line at failure that 

//^'"^^"^^yX corresponds to d^ =: d — ^, the radius 

/ / \ \ of the outer circle being r and t the 

hL -I - - . . .v|( j. ji thickness of the metal, both to be ex- 

\ >v ^ yy pressed in inches. The numerical value 
\^^^J^^/ of d.j, is that required for rupture of the 

material in a hollow circular beam, with 
its full compressive and tensile strength. 

116. Case IV. —Columns that cross-bkeak from com- 
pression. 

The shortest pillar of this class, for a given section and 
material, is that (Fig. 31) in w4iich gn — 6 = d^, or that 
depth of extended area required for equilibrium when a beam 
of this section and material is broken by a transverse load, and 
the longest is that that just deflects sufficiently to allow the re- 
sultant of the applied load, Z, to pass through the fulcrum,/ ; 
this varies with the section, for a given material. 

* From the United States Government Watertown Arsenal Tests for 
1883-84, Senate Ex. Doc. No. 35— 49tU Congress, 1st Session, p. 369. 



STRENGTH OF COLUMNS. 



131 




The load is the siom of the two pressure wedges. In rect- 
angular areas or those that may be divided into rectangular 
areas, the load is the compressed area, multiplied by the crush- 
ing strength of the material. 

Rectangular Pillars. 

The position of the neutral line must 
be computed from either Eq. 25 or 26, 
page 38. Then, in order that the breaking 
load of a given pillar may be deduced 
from the principles applicable to Case lY., 
(5", the deflection, must be equal to or 
greater than ^4, and equal to or less than 
\d. 

Then 

L = M,C, (99) 

from which the required load may be computed, being the sum 
of the two pressure wedges required for rupture. 

Example 46. — Required the breaking load of a Rectangular 
Yellow Pine Column, tested with flat ends, when 

The diameter.. cZ = 5".5, C = 5230 pounds mean of four tests, 
" breadth.. J =5".5, r = 15178 " " "tests, 

" deflection (^ = 0".66, 17 r:^ 0.337. 

The position of the neutral line, computed from Eq. 26, 
page 38, is d^ = 5". 39, and d^ = 0.11, being less than the de- 
flection, the column failed by cross-breaking ; the deflection 
being less than one third of the least diameter, it failed with- 
out cantileverage. 

. • . Z = 5.5 X 5.39 X 5230 = 155043 pounds. 

In a series of experiments to test the strength of yellow pine 
columns with flat ends, conducted on the United States Gov- 



132 



STRENGTH OF BEAMS AND COLUMNS. 



ernment testing macliine, at Watertown Arsenal, 1881-2,* 
thirty columns were broken, whose lengths varied from I = ^1d 
to ^ = 45<;/, which appears to be the limits of cross-hreaking^ 
without leverage for the yellow pine tested. The crushing 
strength obtained from short blocks differed so greatly that a 
msan was not admissible, though two consistent classes can 
be made of the material from these tests. The mean of four 
tests was C = 5230 pounds, from which the required load in 
the above example was computed ; the mean load from three 
tests was 154000 pounds. The mean value of C from three 
other tests was 3600 pounds, from which the computed load 
in the following Table was obtained. The tensile strength was 
obtained on the same machine the previous year, and not 
from the material composing the columns tested. 

The " experimental '' load in the Table is a mean of the 
number of tests given in the last column ; from " failure at 
knots and diagonally," the remainder of the thirty experiments 
mentioned above were rejected. 





Dimensions in Inches. 




Load in 


Pounds. 




I -*• d. 








Deflection in 
Inctiep. 






No. of 














Tests. 




I. 


b. 


d. 




Computed. 


Experiment'] 




27.0 


180 


15.6 


6.6 


.5 to .62 


344822 


409000 


3 


30.8 


180 


12.0 


5.8 


.73 " 1.26 


235000 


250000 


1 


31.2 


240 


9.67 


7.7 


.82 " 1.6 


249253 


281000 


2 


31.2 


180 


15.5 


5.6 


.52 


290718 


344000 


2 


32.8 


180 


5.5 


5.5 


.52 


101574 


129500 


2 


36.0 


180 


12.1 


5.0 


1.3 " 1.8 


202554 


230000 


1 


38.2 


210 


5.5 


5.4 


1.03 '' 1.95 


100584 


97330 


3 


40.0 


180 


11.6 


4.4 


.9 " 1.30 


170800 


126350 


2 


43.0 


320 


9.28 


7.4 


.99 " 2.55 


231516 


199830 


3 


45.0 


240 


5.4 


5.4 


1.28 


98776 


84900 


2 


45.0 


180 


11.35 


4.1 


.95 


166394 


142000 


1 



In the above Table the second and sixth tests carried the 
maximum load with deflections that varied a half an inch in 
amount before failure. 



* Senate Ex. Doc. No. 1 — 47th Congress, 2d Session, p. 321. 



STRENGTH OF COLUMN'S. 



183 



Example 47. — Required the breaking loads of the series 
of White Pine Cohinins tested with flat ends, on the United 
States Government machine, at Watertown Arsenal, 1881-2,* 
when T = 10000 pounds mean of other tests, C — 2500 pounds 
mean of five of these tests. The position of the neutral line 
from Eq. 26 for «/ = 0.25 is 

d^ = 0.92rZ. 

The breaking loads in the following Table were computed 
from Eq. 99, as in Example 46 ; the first pillar of the Table did 
not deflect quite sufficiently to belong to this class of columns, 
but as the error is small, the lower limit is taken to be, as in yel- 
low pine, Z ~ 27<:Z ; the greater limit was not determined by 
the tests. 

The " experimental loads " and the dimensions are the means 
of those given for three tests. 





DiMEN 


SIGNS IN Inches. 




Load in 


Pounds. 










Deflection in 
Inches. 






/-»- d. 














I. 


h. 


d. 




Computed. 


Experimental. 


27 


180 


15.6 


6.66 


.4 to .3 


239400 


227300 


32 


180 


15.6 


5.62 


.37 " .83 


202800 


164100 


32 


240 


9.4 


7.45 


.7 " 1.2 


161200 


170300 


33 


180 


11.3 


5.4 


.6 


165400 


156200 


36 


280 


9.6 


7.69 


.66 " 1.2 


169900 


153300 


40 


180 


11.6 


4.48 


.43 " 1.25 


119700 


130700 


43 


320 


9.3 


7.47 


.7 " 1.67 


159700 


147600 



CiRCULAK Pillars. 

For this Case the deflection, (5^, must be equal to or greater 
than d^^ the depth of the tension area given by the Table be- 
low for the neutral line, when C -^ T = </, for the material, 
and equal to or less than f 6?, the diameter. 

The volume of one pressure may be computed from E(p IT, 



* Senate Ex. Doc. No. 1— 47th Congress, 2d Session. 



134 



STRENGTH OF BEAMS AND COLUMNS. 



page 16, or from computing and tabulating the vohtmes for all 

required values of d^^ when the radius 
is iinitij^ which we will represent by 
fC^ and since the volumes of cylindrical 
wedges are as the squares of their radii, 
we have for the required load, Z, 




L = 2/'y6^. 



(100) 



Table of positions of the neutral line, d^^ and the computed 
values of f for the various required values of </ = C -^ T. 



Q- 


rfc. 


/. 


Q. 


dc. 


f. 


8.0 


0.3984 d 


0.4877 


4.0 


0.4951 d 


0.6578 


7.875 


0.4004 d 


0.4911 


3.866 


0.5000 d 


0.6666 


7.75 


0.4022 d 


0.4942 


8.75 


0.5045 d 


0.6749 


7.625 


0.4041 d 


0.4975 


3.625 


0.5095 d 


0.6838 


7.5 


0.4062 d 


05011 


8.5 


0.5146 d 


0.6947 


7.375 


0.4087 d 


0.5054 


3.875 


0.5200 d 


0.7034 


7.25 


0.4112 d 


0.5097 


3.25 


0.5256 d 


0.7137 


7.125 


0.4137 d 


0.5139 


3.125 


0.5314 d 


0.7289 


7.0 


0.4161 d 


0.5180 


3.0 


0.5375 d 


0.7354 


6.875 


0.4186 d 


0.5221 


2.875 


0.5488 d 


0.7469 


6.75 


0.4209 d 


0.5265 


2.75 


0.5505 d 


0.7691 


6.625 


0.4236 d 


0.5311 


2.625 


0.5574 d 


0.7728 


6.5 


0.4262 d 


0.5855 


2.5 


0.5646 d 


0.7851 


6.375 


0.4289 d 


0.5403 


2.375 


0.5724 d 


0.7993 


6.25 


0.4316 d 


0.5446 


2.25 


0.5804 d 


0.8148 


6.125 


0.4344 d 


0.5496 


2 125 


0.5890 d 


0.8802 


6.0 


0.4373 d 


0.5549 


2.0 


5980 d 


0.8469 


5.875 


0.4402 d 


0.5599 


1.875 


0.6076 d 


0.8641 


5.75 


0.4432 d 


0.5652 


1.75 


0.6178 d 


0.8838 


5.625 


0.4463 d 


0.5606 


1.625 


0.6277 ^ 


0.9042 


5.5 


0.4494 d 


0.5761 


1.5 


0.6404 d 


0.9260 


5.375 


0.4526 d 


0.5819 


1.875 


0.6525 ^ 


0.9496 


5.25 


0.4559 d 


0.5877 


1.25 


0.6666 d 


0.9747 


5.125 


0.4594 d 


0.5987 


1.125 


0.6816 rZ 


1.0039 


5.0 


0.4629 d 


0.6000 


1.0 


0.6976 ^ 


1.0331 


4.875 


0.4665 rZ 


0.6066 


0.875 


0.7162 d 


1.0661 


4.75 


0.4702 d 


0.6182 


0.75 


0.7357 rf 


1.1042 


4.625 


0.4740 rZ 


0.6199 


0.625 


0.7588 d 


1.1466 


4.5 


0.4780 d 


0.6270 


0.5 


0,7843 c^ 


1.1946 


4.375 


0.4821 rf 


0.6348 


0.875 


0.8149 rf 


1.2511 


4.25 


0.4863 rf 


0.6419 


0.25 


d 




4.125 


0.4906 d 


0.6496 1 









STREIS^GTII OF COLUMNS. 135 

Example 48. — Required the breaking load of a Cylindrical 
Cast 'Iron Column, when 

The diameter d = l'M29, O = 96280 lbs. computed in Ex. 3, 
" length... Z=: lO'^O, T= 294:00" by test, 
" deflection d= 0"A, q =3.271. 

The value of / from the Table corresponding to this value 
of 3' is y = 0.715, then from Eq. 100 we have 

Z = 2 X (0.5645)' 0.715 X 96280 = 43884 pounds. 

This column was cast from the iron described in Example 
14, page 82, ' ' Cracks developed on the tension side when 
the deflection reached 0'^4, the load sustained being 40000 
pounds." Four other columns of the same iron were broken, 
w4th deflections varying from 0'^38 to 0''.43, but the loads 
sustained were not given. The theoretical determination of 
the position of the neutral line and the practical is almost 
identical, the theoretical being dr, = 0.538, and the practical 
d^ = O^'AO. 

Hollow Circular Pillars. 

For this Case the deflection, S^ must be equal to or greater 
than drj., the depth of the tension area given by the following 
Table for the neutral line, when q = 
O -i- T for the material, and equal to or 
less than JcZ, the diameter of the outer 
surface of the pillar. 

The volume of one pressure wedge may 
be computed from Eq. 20, page 20, or 
from computing and tabulating the vol- 
umes for all required values of d^, when the radius is unity, 
which we will represent by tfQC, and since they are as their 
radii we have for Z the load, t being the thickness of the 
metal — 

Z = 2rtfoO. (101) 




136 



STRENGTH OF BEAMS ATTD COLUMNS. 



Table of positions of the neutral line, d^., and the computed 
values of f^ for the various required values oi q ^^ C ^ T. 



Q- 


d.- 


U 


g- 


d.- 


U 


8.3195 


0.5000 d 


2.0000 


\ 4.5 


6520 


d 


2.3377 


8.0 


5093 d 


2.0214 


4.375 


0.6590 


d 


2.3528 


7.875 


0.5131 d 


2.0299 


4.25 


6662 


d 


2.3684 


7.75 


0.5170 d 


2.0389 


4.125 


0.6735 


d 


2.3845 


7.625 


0.5209 d 


2.0481 


4.0 


0.6811 


d 


2.4010 


7.5 


0.5249 d 


2.0566 


3.875 


0.6889 


d 


2.4180 


7.375 


0.5290 d 


2.0657 


3.75 


0.6970 


d 


2.4355 


7.25 


5331 d 


2.0752 


3.625 


0.7052 


d 


2.4533 


7.125 


0.5374 d 


2.0847 


3.5 


0.7137 


d 


2.4717 


7.0 


0.5418 d 


2.0945 


3 375 


0.7223 


d 


2.4906 


6.875 


0.5462 d 


2.1044 


3.25 


0.7313 


d 


2.5101 


6.75 


0.5507 d 


2.1144 


3 125 


0.7405 


d 


2.5301 


6.625 


0.5553 d 


2.1248 


3.0 


0.7500 


d 


2.5509 


6.5 


0.5600 d 


2.1351 


2.875 


0.7597 


d 


2.5722 


6.375 


0.5647 d 


2.1457 


2.75 


0.7697 


d 


2.5941 


6.25 


0.5697 d 


2.1568 


2 625 


0.7800 


d 


2.6167 


6.125 


0.5748 d 


2.1680 


2.5 


0.7906 


d 


2.6400 


6.0 


0.5800 d 


2.1790 


2.375 


0.8014 


d 


2.6638 


5.875 


0.5852 d 


2.1910 


2.25 


0.8122 


d 


2.6877 


5.75 


0.5907 d 


2.2035 


2.125 


0.8241 


d 


2.7143 


5.625 


0.5962 ^ 


2.2151 


2.0 


0.8358 


d 


2.7404 


5.5 


0.6018 d 


2.2277 


1.875 


0.8478 


d 


2.7673 


5.375 


0.6076 (^ 


2.2104 


1.75 


0.8600 


d 


2.7950 


5.25 


0.6134 d 


2.2530 


1.625 


0.8724 


d 


2.8234 


5.125 


0.6193 rf 


2.2663 


1.5 


8851 


d 


2.8526 


5.0 


0.6255 6^ 


2.2798 


1.375 


0.8979 


d 


2.8823 


4.875 


0.6314 d 


2.2933 


1.25 


0.9107 


d 


2.9124 


4.75 


0.6384 (Z 


2.3079 


1.125 


0.9233 


d 


2.9425 


4.625 


0.6451 d 


2.3226 


1.0 


0.9360 


d 


2.9733 



Example 49.— Required the Greatest Strength of a Phoenix 
Column, when 

The outer diameter ... .d =^ 8''.00, ^ = 1, 
" thickness of metal, t = 0".35, /„ == 2.9732 for q = 1, 
" deflection. . .' 6 ^ 0''.535, '^ I = T'.O. 

The crushing value of 6^ = 60000 is assumed, then from 
Eq. 101, 

Z =r 2 X 4.0 X 0.35 X 2.9732 X 60000 ^- 499497 pounds. 



STRENGTH OF COLUMNS. 187 

Bj the United States Government Watertown Arsenal 
tests the greatest strength was 468000 pounds."^ 

117. Case V.— Pillaks that cross-break from com- 
pression AND CANTILEVERAGE, 

The shortest pillar of this class, for a given section and 
material, is that length that just deflects sufficiently to allow the 
resultant of the applied load to pass to the right of the fulcrum 
f (Fig. 34). This for a given material varies with the section, 
and the class includes all pillars of longer lengths. The 
amount of direct pressure along the \midfn is equal to the sum 
of the two equal pressure wedges described in Case lY. for 
the given cross-section, and is the same for all deflections, but 
to obtain the breaking load of the pillar this si^n must be di- 
minished by the tensile and compressive strain that arises from 
the load's action as a cantilever. 

C ' = the greatest intensity of the compressive strain arising 
from bending action of the load. 

Rectangular Pillars. 

S — \d ^ the lever-arm of the load, L. 

Then the moment of the applied load will be, from Eq. 23, 
p. 38, 

' r^' -L {?>6-d) 
hd: ' 

Then we will have for the applied load, Z, 

Z = hd,C-hd,C\ 



* Ex. Doc. No. 23, House of Representatives, 46tli Congress, 2d Session, 
p. 278. 



138 STRENGTH OF BEAMS AND COLUMNS. 

from which, hy substituting the above value of C\ we ob- 
tain 

^ ;^3-:r7^- (102) 

When SS becomes equal to or less than d, the depth, the for- 
mula becomes that for rectangular pillars 
of Case lY., the cantilever effect of the 
load having disappeared. 

As before stated, the formulas deduced 
for the strength of columns that fail by 
cross-breaking with cantileverage are iden- 
tical in form with those given by Mr. 
Lewis Gordon, but, unlike his, they give 
exact values and show when the formula does not apply. This 
uncertainty as to the length of pillars to which Gordon's for- 
nmlas did not apply has been the chief objection to their use 
in practice. 

Adapting Gordon's formula for the strength of rectangu- 
lar pillars to our notation, we have 

J. AC 



1 + cI 



From making A the full area of the column, the factor C 
could never be the crushing strength of the material, but is 
always something less than it in value. The second member 
of the denominator makes the formula true for all lengths of 
pillars, which experiment does not confirm, ^nd is of the same 
general form as our formula for the deflection given by Eq. 
93. 

^o successful effort has ever been luade to so modify Gor- 
don's formula that the computed and experimental strength 



STRENGTH OF COLUMNS. 139 

of wooden columns would be the same in value ; and for 
the reason that in all published results of experiments on 
wooden pillars, except for a few on yellow pine made at 
the Watertown Arsenal, they all failed without can ti lev- 
erage of the applied load, and the deflection did not enter 
as a factor to decrease the strength as contemplated by Gor- 
don. 

In the tests of white and yellow pine columns made at the 
Watertown Arsenal, and described in Examples 46 and 47, the 
strength was not decreased from cantileverage. For any 
given rectangular section of these pillars, the strength is the 
same for all lengths from twenty-seven to forty-five times the 
least diameter or least side of the rectangle, which includes 
the lengths of all yellow and white pine pillars that are used 
in structures. 

Example 50. — Required the breaking load of a Rectangu- 
lar Wrought-Iron CJolumn, tested with two V'^ pin ends, 
when 

The diameter d = 3".0, O = 50000 pounds assumed. 

" l)readth h = 3'.0, T = 50000 pounds mean tests. 

" deflection 6 = 1".61, q = 1. 

The position of the neutral line from the Table, page 92, is 
d^ = 0.7Sd; from Eq. 102 we have 

ax 1.61 -3.0 1 + 0.783 ^ 

"^ 2.34 

This example and those in the following Table are from tlie 
series of tests described in Example 44, page 128, as given in 
" Senate Ex. Doc. Ko. 5— 48th Congress, 1st Session," pp. 68 
to 102. 



140 



STRENGTH OF BEAMS AND COLUMNS. 



Dimensions 3" X 3". 







Deflection 


IN Inches. 


Load in 


Pounds. 


I -^d. 


Length in 










Tiif'hpR 












1. mJ.\j Lx\^D , 


From. 


To. 


Computed. 


Experimental. 


28 


84 


0.5 






216000 


30 


90 


0.5 






218975 


32 


96 


0.5 


'I'.ei" 


197666* 


219000 


34 


102 


0.8 


1.61 


197000 


208500 


36 


108 


0.8 


1.80 


178760 


190500 


38 


114 


0.5 


1.80 


178760 


181125 


40 


120 


0.6 


1.9 


168000 


181750 


42 


126 


0.5 


1.67 


188800 


169700 


44 


132 


0.4 


2.30 


131700 


178950 


46 


188 


0.4 


1.98 


155600 


157825 


48 


144 


45 


2.16 


141100 


155790 


50 


150 


0.45 


2.30 


131700 


158065 


52 


156 


0.35 


2.50 


120088 


155000 


54 


162 


0.30 


2.50 


120088 


147750 


56 


168 


0.30 


2.80 


106000 


149875 


58 


174 


0.85 


2.50 


120088 


128775 


60 


180 


0.30 


2.50 


120088 


126875 



When under the greatest load sustained by the above col- 
umns, they "suddenly sprung" to a cross-hreaking deflection 
with cantilever age. The " Deflection from " in the Table is 
that at which the " sudden spring " began, and " Deflection 
to " is that to whicli it sprung / the deflection in the Table 
is the mean of those of two tests, in most cases. '^ 

The " Experimental " load is the mean of two tests, in each 
case, and is that sustained by the column at the beginning of 
the " sudden spring." From the " Computed " loads it will 
be seen that the momentum of the " sudden spring " caused 
the observed deflections to exceed the trne deflection in many 
examples of this series of tests. 

Circular Pillars. 
6 — \d =^ the lever-arm of the load, Z, 
f = the factor in Table, page 134, 
/, = the " " " pages 80, 100 and 112. 



STRENGTH OF COLUMNS. 



141 



From Eq. 76, page 57, the moment of the applied load, Z, 
becomes 

, _ Z{8d - Sd) 



. C = 



r% 



and for the load from the pressure wedges given by Eq. 100 
we have 

L ^ 2rYC - 2rYC'. 



Substituting for C its value given above, we have 



L=. 



1 + 



2/(8()^ — Zd), 
■rf. 



When 8(^ becomes equal to or less than 
Zd the load must be computed from the 
formula for Circular Pillars, Case lY., 
as in this position of the pillar there will J 
be no cantilever effect of the applied 
load to be deducted from the sum of the 
two pressure wedges. 



(103) 




Hollow Circular Pillars. 

d = the outer diameter in inches, 
r = the " radius " " 
{6 — \d) = the lever-arm of the load, Z, 

/J) = the factor given in Table, page 136, 
/„ = the " " " pages 85 and 104. 

t = the thickness of the metal ring in inches. 



From Eq. 91, page 62, the moment of the applied load be- 
comes 



142 STRENGTH OF BEAMS AND COLUMNS. 

_ A^S-d) 
For the load, Z, we have 

Substituting the value of C ^ we obtain 

When ^S is equal to or less than the outer diameter, d, the 
strength of the column must be com- 
puted from Eq. 101, Case lY., as there 
will then be no cantilever effect to de- 
duct. 




Example 51. — Required the Breaking 
Strength of a Phoenix Column tested 
witli flat ends ; C = T = 60000 is the 
assumed strength of the iron. 

Outer diameter d — 8".0, q = C -^ T = 1, 

Thickness of metal, t = 0".85, /o = 2.9732 from Table, page 136, for g = 1^ 
Deflection 5 = 2".47,/e = 2.6806 " " " 104, " q = U 

From Eq. 104 we have 

^ _ 2 X 4 X 0.35 X 2.9732 X 60000 _ 49949 7 ^ 39^426 
~ ^ , 2.9732 (4 X 2.47 - 8) ~ 1+0.26 
"^ 2 X 4 X 2.6806 

This example is taken from the series of experiments^' de- 
scribed in Example 49, page 136 ; the tested breaking load was 
416000 pounds ; the length was 28 feet. 

* Report of the United States Board appointed to test iron and steel, Ex. 
Doc. No. 23, House of Representatives, 46th Congress, 2d Session, page 
270. 



STREIN^GTH OF COLUMNS. 143 

Angle-Iron, Box, Channel, Eye-Beam and Tee Pillars. 

In pillars of the above sections, the distance of the neutral 
line from that side of the pillar that will most likelj be its 
concave side, when broken, must be computed by the rules 
heretofore given for it in beams of the section of the pillar. 
The sum of the two pressure wedges will then be the product 
of the compressed area, A^^ by the crushing strength of the 
material. 

The distance, g, of the centre of gravity of the applied 
load wedge from the point k (Fig. 34), must be determined, 
which is obtained by computing liF, the moment of the load 
wedge with respect to the axis, A', in which F is the greatest 
intensity of the load at ^', it being zero at m., and 7? a factor 
that depends for its value upon the section ; for any given 
dimensions it reduces to a numerical quantity without assign- 
ing any value to F\ and dividing this moment, ^7% by the 
volume of the load wedge, AF -^- 2, in which A is the area 
of the section, we have 



BF--- 



J: 



9 
and 



iF _2B 

^ __L{S-g) 



Then we have 



Substituting the above value of C, we deduce 



. , AAS^-j). (105) 

B^ is a factor of the mom,ent of resistance that the section 
of the pillar offers to the cantilever bending of the load ; for 
a given section it becomes a numerical quantity without assign- 



144 STRENGTH OF BEAMS AND COLUMNS. 

ing any value to C wlien computed by tlie rules given for the 
mouient of resistance of the section when strained in a beam. 
When S becomes equal to or less than g in value the canti- 
lever strain ceases to exist, and the pillar belongs to Case lY., 

.-. L^ A,C. (106) 

When d becomes less than r/^, the depth of the extended 
area required for rupture, the pillar belongs to Case III. 

in which d^ gives the position of the neutral line of rupture, 
and ^ the ratio of the compressive strain that will be required 
to hold in equilibrium the tensile strain developed by the 
bending of the pillar as a cantilever. 

Equation 105 is the general formula for the strength of 
pillars of all lengths, sections and material of which the 
formulas heretofore deduced in this chapter are only the forms 
it will assume for special cases. The denominator of the 
second member of the formula must never be less than unity. 



CHAPTER YIII. 

COMBINED BEAMS AND COLUMNS. 

118^ General Statement. In roof -trusses, cranes, 
derricks, platforms supported by cantilevers, trussed beams and 
other structures, there is used a class of pieces of material that, 
from the manner in which they are loaded, do not belong ex- 
clusively to either horizontal beams or columns, but partake of 
the nature of both, in the manner in which they support the 
load to which they are subjected. 

The theory of the transverse strength of these Combined 
Beams and Columns gives the solution of the general problem 
of the transverse strength of all beams, without regard to the 
special angle that the axis of the beam makes with the line of 
direction of the loading and supporting forces. Horizontal 
beams acting under vertical loads and columns are only special 
cases of tlie general problem, in which certain factors, that 
cause the strength of the same piece of material to vary with 
its angle of inclination to the horizon, disappear, from the 
general rale for these cases, by becoming zero in value. 
But on account of their great importance, we have, in the 
preceding chapters, deduced separately the principles and rules 
from which the strength of these special cases may be com- 
puted. 

Should one end of a horizontal beam, such as Ije (Fig. 44), 
be fixed in a vertical wall, and a load be attached to its free 
end, not suspended, as in the figure, none of this load will di- 
rectly compress or rest upon the cross-section of tlie beam, he. 
Now let the wall be revolved around the point 7^, to a horizon- 
tal position, thus bringing the beam, he^ to the vertical, then the 



146 STRENGTH OF BEAMS AND COLUMNS. 

entire load will rest upon or directly compress the cross-section 
of the beam, now converted into a pillar ; the load has thus been 
gradually converted from a.non-compressing to a compressing 
load with its full weight. The bending moment of the attached 
load is greatest when the beam is in the horizontal position, and 
it gradually diminishes as its lever-arm, s^ becomes less in value 
with the revolution of the wall, and becomes zero in value 
when the beam occupies the vertical position. On the other 
hand, should the wall be revolved to the horizontal position 
around the point P^ the bending moment will gradually de- 
crease and become zero, while the tensile strain w^ill increase 
from zero to that of the full w^eight of the load, when the 
beam becomes vertical. 

From the above illustration the origin of the special cases 
of horizontal beams and colicmns is apparent, and the reason 
for the special rules for their strength. A similar illustration 
could be deduced from a beam supported at both ends, by con- 
ceiving it to occupy all positions from the horizontal to the 
vertical. 

In order to deduce the relation that exists between the 
applied load and the resistance of the combined beam and 
column at the instant of rupture^ it is assumed., in the analy- 
sis, that they only deflect enough to admit of their cross-hr cak- 
ing as a column, without cantileverage of the applied load, as 
in Class lY., Chapter YII. In very long beams, how^ever, the 
compression resulting from pressure applied to its ends will 
act with the leverage explained in Case Y., page 137, and the 
strength must be computed from tiie formulas there given, but 
the compression resulting from the transverse bending of the 
load will be the same in each case, and will compress the beam 
without cantileverage. 

119. Notation. In addition to the notation heretofore 
used and defined in Art. 35, page 37, the following will be 
used in this Chapter : 



COMBINED BEAMS AND COLUMNS. 147 

(7 = 0'-{- C" ^=^ tlie greatest intensity of the compressive strain 

in pounds per square inch, 
C = the greatest intensity of tlie compressive strain in pounds 
per square inch, arising from the bending component 
of the load, 
C"^= the greatest intensity of the compressive strain in pounds 
per square inch, arising from the compressing compo- 
nent of the apphed load, 
Zb = the bending component of the applied load in pounds, 
Zc == the compressing component of the applied load in pounds, 
I = the unsupported length of the beam in inches, 
s = the span, the horizontal distance between the supports 

in inches, 
h — the difference between the heights of the ends of the 

beam in inches, 
a = the angle that the beam makes with the horizon. 

In many of the different methods of loading and support- 
ing beams given in this Chapter, the greatest resulting com- 
pressive strain at any section is the sinn of two or more dis- 
tinct pressures, which will be represented by the letter O, with 
corresponding accents, such as C\ C\ C"\ C"'\ etc. 

Inclined Beams. 

120. General Conditions. The effect produced by 
a load when applied to this class of beams will manifest itself 
in two distinctly different ways ; each separate effect must be 
computed, and their sum will be the total effect produced by 
the load upon the beam. This is accomplished by decompos- 
ing the applied load into two components^ by the well-known 
theorem of the parallelogram of forces. One component, Z^, 
must be at right angles to the axis of the beam, and the other, 
Zc, parallel with it ; the first component will bend the piece as 
a heam, while the second will compress it as a column. 

The parallelogram of forces for inclined beams is a rect- 



148 



STRENGTH OF BEAMS AND COLUMNS. 



angle, and the relation between the components and the load 
is obtained from that of the three sides in a right-angle tri- 
angle, in which 



COS. a =^ — 



h 

sin. a = — 

V 



, h 

tan. a=^ -. 

s 



121. Inclined Beam Fixed and Supported at 
One End. This method of " fixing " and loading beams is 
illustrated by two different positions of the beam, he^ in Fig. 
44. Two Cases will be considered. 

Case I. — When the load is applied at the fi^ee end of 
the heam. 

Let the right-angle triangle, eot (Fig. 44), represent the 
parallelogram of forces, in which et represents the applied 




load, drawn to any given scale, then ot will be the hending and 
oe the compressing component of the applied load, Z, from 

which 

ot =: et COS. a and oe = et sin. a, 

Ls 

T' 



L. 



(108) 



COMBINED BEAMS AND COLUMNS. 149 

and 

A = ~-^ (109) 

The bending moment of the component, Z^, its lever-arm 
being /, will be 

Bending Moment, L^ = -^Xl = Ls. (HO) 

Having decomposed the applied load, Z, into two compo- 
nents, one perpendicular and the other parallel to the inclined 
beam, we can now ascertain the effect that will be produced by 
the original load upon it, by a combination of the methods 
used to compute the effect produced upon horizontal beams 
and columns by vertical applied loads. 

Rectangular Beams. 

The transverse strength of a rectangular beam loaded and 
fixed as in this Case will now be deduced from the foregoing 
formulas. 

The bending moment from Eq. 110 must be made equal to 
the moment of resistance from Eq. 23, page 38, in which 

r.Ls='^, (111) 



3 
ZLs 



d. 



hd,C'. (112) 



The compressing component, Z^, from Eq. 109, must be re- 
sisted by an equal compression produced in the section at the 
face of the wall, as given by Eq. 99, page 131, in which 



C = C" and L = ^^ 



^=:hd,G", (113) 



150 STRENGTH OF BEAMS AND COLUMNS. 

Adding Eqs. 112 and 113 we have, by making C ' -{- C " ^=. 
C^ the crushing strength of the material, 

from which 

L = _J^^_. (115) 

When the beam is horizontal, h^^O and I = s, the formula 
reduces under this hypothesis to that given in Eq. 27, page 38, 
in which m = 1. In a vertical beam, ^ = and I = h, the 
fornmla reduces to that given for the strength of rectangular 
columns, Eq. 99. 

To Design a Rectangidar Seam. 

The length I and height h will be controlled by the position 
in which the beam is to be used. A convenient depth, d, 
must then be assumed, and the value of d^ computed from 
Eq. 26, page 38 ; then from Eq. 114 we obtain 

from which the required breadth, h, of a beam that will 
break with a given load, Z, will be obtained by giving to O 
the value of the crushing strength of the material of which 
the beam is to be constructed. 

Case II. — When the Load is uniformly distributed 

OVER THE unsupported LENGTH OF THE BEAM. 

The resultant of the applied load will pass through the 
middle of the length of the beam and the triangle, eot (Fig. 
44), will, as in Case I., give the relation between the load and 
its two components ; hence 

A = ^, (in) 

Ze = ^- (118) 



COMBINED BEAMS AND COLUMNS. 151 

The bending moment of the component, Z^, will be, 

Bending Moment =^ — - x ^- — --5 (^l^) 

its lever-arm being Z -^ 2. 



Rectangular Beams. 

The transverse strength of a rectangular beam loaded and 
fixed as in this Case may be obtained by the same process 
heretofore used in Case I., 

• /- ^i^^ (120-, 

To Design a Rectangular Beain. 

Assume a depth, r7, and compute J^ from Eq. 26, page 38, 
then from Eq. 120 we have 

, _ {Zh + 2r4A) L 



133. Inclined Beam, supported at one end 
and stayed or heUl in position at the otlier with- 
out vertical support. This class of Inclined Beams is 
illustrated in Figs. 45, 46 and 47. Five different cases of 
loading will be considered. 

Case I. — When the Inclined Beam is loaded at its 

STAYED END. 

The effect produced by the load, Z, upon the beam, he (Fig. 
45), is simply to compress it as a colunm, the load being held 
in equilibrium by a pnll along the tie ec^ and a thrust along 
the inclined beam. 

In the right-angle triangle, toe^ let et represent the load 



152 



STRENGTH OF BEAMS AND COLUMNS. 




drawn to any given scale, then to gives the pull and oe the 
thrust. 



et te 

01 = -: • 5 oe = -. — 



tan. a 



sin. a 



Ls 
The pull = -J-, 



and 



The thrust L, 



n 

h 



(122) 



(123) 



from which the compressing effect of the applied load, Z, 
can be ascertained. 

Rectangular Beams. 

The direct compression given bj Eq. 123 must be made 
equal to the resistance offered bj the section of the beam as a 
column, then from Eq. 99, page 131, 

h ' 



L = 



hhd.C 
I 



(124) 



COMBINED BEAMS AND COLUMNS. 



153 



Case II, — When the Inclined Beam is loaded at its 

MIDDLE. 

Let he (Fig. 46) represent the beam loaded at its middle, t^ 
witli the load, Z, then in the right-angle triangle, toe^ et rep- 




resents the load drawn to any scale, ot^ perpendienlar to the 
beam, hc^ the hending component, Zg, and ot parallel to he, the 
directly compressing component, L^. 



_ Lh 
A- -J- 



(125) 



(126) 



The hending moment produced by the component, Zg, is 
identical with that produced by conceiving the beam to be 
" fixed " at its middle, t, and loaded at the free end with 

. • . Bending Moment, Zb = — y X ^ == ~T~' (1^^) 



In order that the loading and supporting forces shall be in 
equilihrium, one half of the hending component, L^, must be 
supported at each end of the beam h and c. The half of L^ 
at G cannot be directly supported at that point, but must be 



154 STRENGTH OF BEAMS AND COLUMNS. 

carried to the ground bj some means. Tlie method of stay- 
ing the end of the beam, he, represented in the hgure is tliat 
used in roof -trusses ; hence the load, Z3 -^ 2 at c', must be de- 
composed into two components, one in the direction of each 
beam or rafter ; if the angle, hck, is a right angle the entire 
load, Zb -^ 2, compresses the rafter, ch, as a pillar, thus reach- 
ing the support, Jc. 

In roof-trusses the inclined beams, he and ck, are usually 
loaded in the same manner, and eaeJi beam will, therefore, 
carry to e a component, L^ -^ 2, of its load for support, Avhich 
will be equivalent to the component, L^ -^ 2, compressing 
each rafter to which it is applied, but much increased in 
amount from the manner in which it is converted from a load 
that is perpendicular to the beam into a compressing load 
that is parallel to its axis. 

In the two equal triangles, cmn, let em in each represent the 
load, Ljj -^ 2, €)i the equal components that the rafters exchange 
w^ith each other, and mn the equal components that the 
loads, Z3 -^ 2, produce that strain the rafters to which they 
are applied. When the angle, 5c^, is greater than 90° the 
sum of the components {inn -f ei%) will be greater than L^ -.- 2, 
when hek = 90°, {?nn + en) = Z^ -^ 2, nm becoming Izero. 
When hek is less than 90° the component, mn, is a tensile- 
strain ; the total compression {cm — m?i) wdll then be less than 
Z«-2. 

In roof -trusses the angle, hek, is generally greater than 90°, 
and the total compression upon each rafter resulting from the 
component of the load that it carries to c is {mn -\- en). 

mn = me . tan. (90° — 2<^), 
7n.e 



en = 



eos. (90° - 2a)' 



Addlns: and substitutino: me — -—r, we have 



Ls^ 

21 



COMBINED BEAMS AND COLUMNS. 155 

(™. + ..) = ^ = |(.«».(90»-2.)+— ^J.-^,). (128) 
The factor, (?, in this equation, for convenience is 

. = .«n,.(90»-2.) + --^_^^. (129) 

Wlien the angle a ^ 45°, tan. (90°— 2a) := 0, — 4 — ^--^ = 1, 

COS. ( t/U JiCtj 

and the amount of compression will become — ? =: —^ , the 

angle, hck^ being a right-angle. 

The total compression produced by the single load, Z, at the 
middle of each rafter will be the sura of the three distinct parts, 
represented by Eqs. 126, 127 and 128. 

Rectangulak Beams. 

The amount of direct compression from Eqs. 126 and 128 
must be equal to the resistance of the beam as a column, from 
Eq. 99, page 131, in which C — C and C— C" respectively. 

.-. ^=:hd,C\ (130) 





and 



:^^ := M,C'". (131) 



The hending Tuionient from Eq. 127 must be equal to the 
moment of resistance of the section as a beam from Eq. 23, 
page 38, in which C ^ G" . 

• • -T'"~3~' ^^^^^ 

.-. ?I:'L = hd,0'\ (133) 

4:d^ 

Adding these separate components, as given by Eqs. 130, 



156 STRPJNGTH OF BEAMS AND COLUMNS. 

131 and 133, and making C'+ C" + C" = C, the greatest 
compressive strength of the material, we have 

.-. L = iii^ Q35X 

35^ + (4A+2c6')6^e 

from which the required breaking load may be computed. 
Case III, — When the Load is uniformly distributed 

OVER THE LENGTH OF THE INCLINED BeAM. 

The bending moment and the component L^ will be identi- 
cal with those produced by conceiving one half of the total 
load, Z, to be concentrated at its middle section, t (Fig. 46) ; 
then, from Eqs. 125, 126 and 127, Case 11., by making 
Z = Z -i- 2, we have 

L. = §, (136) 

A = ~, (137) 

Ls 

Bending Moment, L^^^ ^-' (1^^) 

o 

The bending moment is produced by a component of the 
load, Z, that is perpendicular to the length of the beam, and 
uniformly distributed over its length, he. In order that equili- 
brium shall exist, one half of this uniformly distributed com- 

ponent, — -^ must be supported at each end of the beam, and as 

it cannot be directly supported at the stayed end, c, it must be 
carried to the supports h and k, as described in Case II., Eq. 
128. 

The method of analysis of the strains in a simple roof-truss, 
given in this and the preceding Case, differs from that usually 



COMBINED BEAMS AND COLUMNS. 157 

pursued by writers. One half of the load on each rafter is 
usually considered to be supported by h and h^ and the other 
half of eacli rafter load to be supported at c, by reacting against 
each other. The transverse strength is then computed as if the 
rafter were a horizontal beam, which is equivalent to saying, 
that for a given material, section and span of roof, as in Fig. 
46, the transverse strength would be the same for the infinite 
number of roof -trusses that could be constructed between these 
points of support by varying t[\Q jpitch or angle that the rafter 
makes with the horizon, which, of course, is not the case. 

Rectangular Beams. 

From Eqs. 128, 137 and 138, ,we obtain, by a process similar 
to that used in Case III., 

Z (--\- - 4- — _) = hd,C, (139) 

. • . Z = Sj^^ , Q40) 

3,9^+(4A + 2t'^)^e ^ 

Example 52. — Required the uniformly distributed breaking 
load of a White Pine Rafter, as in Fig. 46, when 

The span 5 = 15'. 0, 6^ = 5000 pounds mean of tests, 

" length Z = 16'. 8, Z = 10000 " " " " 

" rise A = 7'. 5, ^ = 0.5, 

" depth d= 9''.0, d^ = OMSd, from Table, page 108, 

" breadth... &= 5^0, ^ ^ 26°.34'. 



From Eq. 129, <? = 2, then, from Eq. 140, 

^ _ 8 X 201.6 X (7.8 1 2X^X^000 ^ ^^^^ 

3X180X201.6+(4X 90+2x2x1 8). 868X9 

Tliis Example is taken from Trautwine's "Engineers 
Pocket-Book," and was designed to sustain an uniformly dis- 
tributed load of 8000 pounds, with a factor of safety of three. 



158 



STRENGTH OF BEAMS AND COLUMNS. 



Case IV. — When the Inclined Beam is Loaded at its 

MIDDLE, AND WITH AN ADDITIONAL LoAD AT ITS STAYED END. 

The bending moment and the compression produced by the 
load, Z, applied at t^ the middle of the beam, he (Fig. 47), will 




be obtained from Eqs. 126, 127, and 128 of Case II., the two 
conditions of loading being the same. 

One half of the additional load, Z^, suspended from the 
stayed end, <?, is supported by each support, l and h ; the com- 
pression produced upon the inclined beam as a column has 
been deduced in Case I., Eq. 123; tlie horizontal components, 
ZTand F^ acting in opposite directions, neutralize each other. 

From Eqs. 122 and 123, page 152, we have, by making 



L.s 



Horizontal corrvponent^ II = -^5 



ThM^ Thrust, L^ = 



2A' 



(141 



(142) 



Rectangular Beams. 

The amount of direct compression from Eqs. 126, 128 and 
142 must be made equal to the resistance oi the beam as a col- 
umn, as given in Eq. 99, 



Lh_ 
I 


^hd,C\ 


Lcs _ 
2/ 


:-ld,C"', 


LJ _ 
'2h 


= M,G"'' 



Ls 


m:c" 


4 


- 3 ' 


4fZ^ 


^hd,C", 



COMBINED BEAMS AND COLUMNS. 159 

(143) 
(144) 

(145) 

The hending Tnovient from Eq. 127 must be equal to the 
moment of resistance of the section of a beam from Eq. 23, 
page 38, in which 0^ C", 

(146) 

(147) 

Adding Eqs. 143, 144, 145 and 147, and making 

C + C" + C""+ C"" = 0, the crushing strength of the ma- 
terial, we have 

A(3.^^-j-2r/, (2/^ + 6'^))' ^ ^ 

from which the centre breaking load of the beam may be com- 
puted, in addition to that of Z^ suspended directly from its 
stayed end. 

Case V. — When the Load is uniformly distributed 

OVER the length OF THE INCLINED BeAM, AND AN ADDITIONAL 

Load, Z^, applied to its stayed end. 

This Case is Cases III. and lY. combined, therefore Eqs. 
128, 137, 138 and 142 will be the formulas required for 
computing the transverse strength of the beam. 



160 STRENGTH OF BEAMS AND COLUMNS. 

Hectangular Beams. 

The direct compression from Eqs. 128, 137 and 142 mnst 
be equal to the resistance of the beam, as a column, from Eq. 
99, page 131. 

If = M.C, (150) 

^^ = ^<^G"'^ (1-51) 

^ = M^G"". (152) 

The hending moment from Eq. 138 must be equal to the 
moment of resistance of tlie section as a beam from Eq. 23, 
page 38, in which C =^ C\ 

(153) 

(154) 
Adding Eqs. 150, 151 and 152, we obtain by making 

^ (fe + 1 + i) + S = ''■'''' (^^^) 

. ^ _ 4ld,(2hbd,C-Z,l) 
• k{3sl+4:d,{h + cs))' ^ ^ 

123. Inclined Beam Supported vertically at 
both ends and loaded at its niiddleo 

Let Fig. 49 represent the beam, supports and the load, Z. 
In the triangle, toe, let eo, to any scale represent the load, Z, 
then the component, of, at right angles to the beam, will bend 
it and et parallel to the beam will compress it as a column. 



Ls 


m:g" 


8' ~ 


3 ' 


ZLs 


M,C". 



COMBINED BEAMS AND COLUMNS. 



161 




ot 



Fig. 49. 

eo COS. a and ot = eo sin, a, 
Ls 



and 



.-.Z. 



L. 



Lh 
I 



(157) 



(158) 



One lialf of the bending component, Z^, will be snpported by 
tlie walls nnder each end of the beam ; the bending moment 

will therefore be 

Ls I Ls 
Bending Moment^ Zj, = -r^ X ^ = -^ * (159) 

Having decomposed the load into two components, one 
perpendicular and the other parallel to the axis of the inclined 
beam, we can compute its effect upon a beam of any section 
and material. 

Load Sustained by the Supports. 

Each of the walls, supporting the ends of the inclined beam, 
must resist a thrust of one half of the bending component, Zg, 
which tends to overturn them. Decomposing this thrust at 
each support into a vertical and horizontal component, the 
first will be the proportion of the centre load, Z, that is sus- 
tained by the higher support. The lower support also sus- 



162 STRENGTH OF BEAMS AND COLUMNS. 

tains an equal vertical component, and in addition a vertical 
component of the compressing component of the applied 

load, L. 

The horizontal thrnst against the higher support will be the 
horizontal component of one half of Z^ ; that at the lower 
support will be the sum of the two horizontal components 
arising from decomposing L^ and Zj, ^ 2 at the lower support. 

From the above we obtain, 

U 

Vertical load on higher support = ^ • cos. a. 

Vertical load on lower support — j (^^^'^- ^ + ^^ '^'^'*^' ^)- 

When the beam becomes horizontal the angle a — o, s — /, 
COS. a — \ and the sin. a — o\ substituting the values in 
the above formulas, we find that each wall supports one half 
of the load, Z, as in Case III., page 4. 

Rectangular Beams. 

The direct compression from Eq. 158 must be equal to the 
resistance of the beam as a column from Eq. 99, page 181. 

^ = Id^C. (160) 

The lending motnent, from Eq. 159, must be equal to the 
moment of resistance of the beam, from Eq. 23, page 38. 

Ls __ hd^C" /.^.x 

4 " 3 ' ^ ^ 

,'.^^ = hdj)\ (162) 

4:dc 

Adding Eqs. 160 and 161 we have, by making (■' + 6'" = G, 
the compressive strength of the material. 



COMBINED BEAMS AND COLUMNS. 163 






-,} = hdA (163) 

from wliicli the centre breaking load of any rectangular 
inclined beam supported at both ends may be computed. 

124. Inclined Beams supported at both ends 
and the load viniforinly distributed over its 
length. 

The bending moment and compression are identical with that 
produced by conceiving the beam to be " fixed" at its middle 
and loaded on its free ends with an uniformly distributed load, 
Z -^ 2, as in Case II., page 150, L being equal to Z ^ 2 

^ Ls J Lli 

X3 _ ^ and Ze = -^ • 

The bending moment will be. 

Bending Moment^ L^ = ^ y^ — z=: —- • 

From these components the effect of the uniformly distrib- 
uted load on an inclined beam of any section and material may 
be computed as in Art. 123. 

Load Sustained by the Support. 

For an uniformly distributed load, Z, the amount of vertical 
pressure that will be sustained by each wall will be obtained 
from the formulas before deduced for a concentrated load, Z, 
at the middle of the span, tlie proportion of the total load 
sustained by the walls being the same in each case of loadiiig. 

The method usually given by writers for determining the 
load supported by each wall, and the bending moment of the 
load applied to an inclined beam, is to consider one half of the 



164 STRENGTH OF BEAMS AND COLUMNS. 

total load to be sustained by eacb wall and the bending mo- 
ment to be the same as if the beam were horizontal, the span 
nsed in the com]3utation being the horizontal distance, .y, 
between the supports. This method of analysis is manifestly 
incorrect, as it would require that an inclined beam of a given 
scantliiig should have a transverse strength that would neither 
be varied by its length nor the angle that it makes with the 
horizon, provided its horizontal span remained the same. Or 
that an inclined beam that is infinitely long, and a horizontal 
beam whose span is the distance between the supports of the 
inclined beam, will have the same transverse strength, pro- 
vided the cross-section and material are the same in each beam. 

RectanCtUlar Beams. 
L^ _ hd:C" . ZLs , 

-^ - —r- ■ ■M = ^"''^' ■ (^'^"' 

Adding Eqs. 165 and 166, we have, by making C + C" = C. 

Trussed Beams. 

125. General Conditions. In trussed beams and 
roof -trusses, beams are frequently subjected to both tran-s- 
rerm and longitudinal strains either of compression or tension, 
thus acting in the double capacity of a horizontal beam and a 
vertical column. 

Rectangular Beams. 

Let Fig. 1-S represent such a rectanguhir beam loaded 
transversely with a total load, />, and longitudinally with 
either a compressive or tensile load, L^. 



COMBINED BEAMS AND COLUMNS. 



165 




From Eq. 27, page 38, in which 6^ = C\ we have 



L = 



^Ls 



Zs 



tncL 



hcLC\ 



(168) 
(169) 



And from Eq. 99, page 131, we have 

L, = M,C'\ (170) 

Adding Eqs. 169 and 170, we have, bv making C'-\-C"^^C^ 



tncL 



+ A = hd,a 



(171) 



From which either Z, Zc, or h maj be computed when the 
other two are known. « 

OS 



z.= 



h = 



mM;C - ZLs 



(172) 
(173) 

(174) 



Hollow Circular Beams. 

The relation between the transverse load, Z, and the result- 
ing moment of resistmice of the beam is given by Eq. 91, 
page 62, in which C = C. 



166 STRENGTH OF BEAMS AND COLUMNS. 

/c =: the factor from the Table, pages 81 and 104, 
/; = the " " " " page 136. 



X = mrtjcC — ^L±i/-f — 

• 2/„Z 



mr/^ 



(175) 
= S/Y/^^'^ (176) 



The compression from the longitudinal strain, Zc, must be 
equal to the resistance of the beam as a column from Eq. 101, 
page 136, in which C = C" . 

L, = ^rtf,C". ,(177) 

Adding Eqs. 176 and 177 we have 

+ Ze = ^rtf.C (178) 



2/oZ 



. . X- 2^. - • (l^y) 

The compressing load, Z^ in the foregoing formulas, in 
practice, generally results from the transverse load, Z, its 
effect being transmitted to the ends of the beam by a vertical 
post placed under the centre of the beam and connected with 
the ends by inclined truss-rods. For a single load, Z, at the 
middle it is usually the practice to consider one half of it as 
producing a tensile strain, on each of the inclined truss-rods ; 
this is only strictly correct when the beam is cut into two 
pieces at its centre. When the beam is continuous it can only 
strain the vertical post after it l)egins to deflect ; therefore the 
load that would have caused the untrussed beam to deflect 
from the horizontal position could bring no strain upon the 
post. 



I ]sr D E X. 









PAGE 


Beams. 


See Transverse Strength, Moment of Resistance and Neutral 




Line 








Beams, 


Cast-Iron, 




64 


<( 


Wrought- Iron, 




87 


< I 


Steel, 




87 


i t 


Timber, 




106 


(( 


Horizontal, 




3 


" 


Inclined, 




148 


" 


Trussed, 




164 


<< 


Elastic Limit, 




34 


" 


and Columns, Combined, 




145 


Bendin 


g Moments, 




3 


" 


General Formula, 




6 


( < 


Moment and Moment of Resistance, .... 




22 


Coefficients of Strength, 




33 


Columns, Box Section, 




163 


" 


Classed,, 




123 


< ( 


Deflection of, 




121 


< ( 


Experiments, 




122 


I i 


End Connections, 




124 


" 


Failure of, without Deflection, 




125 


<< 


" with " 




127 


" 


" without Cantileverage, 


, 


130 


' ' 


" with 




137 


" 


General Conditions of Failure, 




114 


< ( 


Gordon's Formula, 




138 


< c 


Pin End, 




118 


" 


Resistance of, 




119 


< < 


Rectangular Sections, 128, 


131, 


137 


" 


Circular Sections, 129, 


133, 


140 


<< 


Hollow Circular Sections, 130, 


135, 


141 


<( 


Angle-Iron, Eye-Beam, Channel, etc.. 




143 


Combined Beams and Columns, 




145 



168 I^^DEX. 

PAGE 

Crushing Strength, To Compute 39, 58, 126 

of Cast-Iron, 64 

To Compute, 67 

" Wrought- Iron, 87 

To Compute, .... 88 

Designing a Rectangular Beam, 39 

" a Hodgkinson " 44 

a Double T " 48 

a Box " 48 

an Inverted T "......• • 53 

a Double T " o3 

Elastic Limits Defined, 33 

" Limit of Cast-Iron, 65 

" *' Wrought-Iron, 87 

" " Steel, 88 

" " Materials, 33 

Equilibrium, 3 

" Moment of Resistance and Bending Moment, . . 23 

Factors of Safety, 34 

Forces Defined, 1 

" Uniform Intensity, • • 2 

" Uniformly Varying Intensity, 2 

Gordon's Formula, 1^38 

Inclined Beams Supported at One End, 148 

" Both Ends, 160 

Load, • • • 2 

Moment of Resistance Defined, 21 

" Box Sections, General Formula, ... 47 

" Circular Sections," " . . .55 

Double T Section,'' " ... 47 

Hodgkinson " " " ... 41 

" " Hollow Circular Sections, General Formula, 60 

'' " " Eye-Beam Section, General Formula, . . 47 

Inverted T, " " . . 52 

*' " Rectangular Sections, " *' . . .37 

" " Concentrated Parallel Forces, 3 

" " Uniformly Varying Forces, 8 

Neutral Line Defined, 21 

" " Box Sections, General Formula, 48 

" " Cast-Iron Sections, 78 

" " " Wrought-Iron and Steel Sections, .... 96 

" " Circular Sections, 56 

" •' " Cast-Iron Sections, 80 



INDEX. 



169 



Neutral Line Circular Cast-iron Sections, Table of, 
" " " "Wrouglit-Iron and Steel Sections, 

" " " " Table of, 

" " " Wooden Sections, 

Table of, . 
" Double T Section, General Formula, 

" " Wrouglit-Iron and Steel Sections, 
" " " " Cast-Iron Sections, ..... 

" " Eye-Beam Sections, 

" " Hodg'kinson Section, General Formula, . 

" " " Cast-Iron Section, .... 

" " Wrought-Iron Section, 

" Hollow Circular Sections, 

" Cast-Iron Sections, 

Table of, . 
" " " " Wrought-Iron Sections, 

Table of , 
" " Inverted T, General Formula, .... 

" " Movement of, 

" " " " from Experiments, .... 

" Rectangular Section, General Formula, 
" " " Cast-Iron Sections, .... 

Table of, . 
" " " Wrought-Iron and Steel Sections, 

" " " Tableof, 

Wooden Sections, Table of, 

" " Position of, 

" "of Rupture in a Rectangular Section, 

" Transverse Elastic Limit, 
Notation for Rectangular Sections, 
" " Hodgkinson Beam, 

" Double T and Box Beams, 
" " Inverted T and Double Inverted T, 

" Circular Beams 

" " Hollow Circular Beams, 

" " Columns, ..... 

" " Inclined Beams, .... 

" " Uniformly Varying Forces, 

Pin End Columns, ...... 

Relative Transverse Strength of Beams, 

" Strength of Square and Circular Beams, 

Cast- Iron Beams 
Timber 



I'AGE 

80 

100 

100 

112 

112 

48 

96 

78 

96 

42 

75 

96 

61 

85 

85 

104 

104 

58 

26 

82 

, 38 

' 71 

71 

92 

92 

108 

24 

28 

24 

37 

41 

47 

52 

55 

60 

121 

146 

8 

118 

35 

59 

83 

113 



170 



INDEX. 



8, 16 



89 



Relative Value of the Crushing and Tensile Strength, . 

Resultant Defined 

. " Uniformly Varying Forces, 

Strain, 

Stress, 

Tensile Strength. To Compute, 

of Cast-Iron, 

«* «' " " To Compute, 

" " " Wrought-Iron and Steel, . '. . . 

" '< " " To Compute, 

" " " Timber, 

«« " " " To Compute, 

Transverse Strength, General Conditions, .... 

Box Beams, General Formulas, . 
" Cast Iron Beams, .... 
" Wrought-Iron and Steel Beams, 

" Cast-iron Beams, 

Circular Beams, General Formula, . 
Cast- Iron Beams, 
Wrought-Iron and -Steel Beams, 
^" " " Wooden Beams, 

Double T Beam, General Formula, . 
'< " " " Wrought-Iron and Steel Beams, 

" " " Cast- Iron Beams, 

Eye-Beams, Wrought Iron and Steel, . 
Hodgkinson Beam, General Formula, 
'« " " Cast-iron Beam, . 

Wrought-Iron and Steel Beams, 
" Hollow Circular Beams, General Formula, 

'* " " Cast-iron Beams, 

" " " Wrought Iron and Steel Beams 

Inverted T, General Formula, . 
Rectangular Beams, General Formula. . 
Cast-Iron Beams, . 
" " " Wrought-Iron and Steel Beams, 

" " " Wooden Beams, . 

Steel Beams, 

" " Wrought-Iron Beams, .... 

" " Timber Beams, 

Trussed Beams, 

Working Load, 



28 
2 

19 

1 

1 

, 58 

64 

67 

87 

90 

106 

106 
21 
48 
78 
96 
65 
56 
81 

101 

113 
48 
96 
78 
96 
44 
75 
96 
61 
86 

104 
53 
38 
71 
92 

108 
92 
91 

106 

164 
34 



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